Learning Objectives

After going through unit, you would be able to understand and appreciate the following:

• The importance of experiments in science.

• The ‘open-ended’ and ‘self-corrective’ nature of science; the role of experiemnts in having this type of nature for science.

• Getting familiar with the basic approach and observations of a few basic experiments.

• Basic devices, like the vernier callipers and the screw gauge and their role in improving the precision of measurements.

• The simple pendulum; its role in providing an objective method for mesurement of time.

• The principle of moments, its use in providing a simple objective method for measurement of mass.

• The concept of ’elasticity’ and Hook’s law.

• The meaning, and method of measurement, of the Young’s modulus for a material.

• The concept of ‘surface tension’ of a liquid; its importance in daily life.

• The reason for rise of a liquid, in a capillary tube, in an ‘apparent defiance’ of the force of gravity.

• The use of the capillary rise method for finding the surface tension of a liquid.

• The concept of ‘viscosity’ of liquid; relative comparison of the viscosities of different liqiuds through their ‘coefficient of viscosity’.

• The concept of ’terminal velocity’; the role of the force of viscosity in making a freeely falling object acquire its ’terminal velocity’.

• The use of ‘measurement of terminal velocity’ to find the coefficient of viscosity.

• The (non-linear) nature, of the fall (with time) of the termperature of a hot body.

• Plotting of cooling curve.

• The concept of ‘resonance’.

• The ‘resonance positions’of an air column when set into vibrations by a tuning fork.

• The use of the ‘resonance tube’to find the speed of sound, in air, at room temperature.

• The concept of ‘specific heat capacity’ of a given substance.

• The basic principle of exchange of ‘heat energy’, or transfer of heat, between different objects.

• Theory of the ‘method of mixtures’ in finding the ‘specific heat capacity of a given (i) solid (ii) liquid.

• Understanding the principle of the Wheatstone Bridge; its use in the meter bridge for finding the resistance of a wire; and hence the resistivity of its material.

• Understanding Ohm’s law and its use in finding the resistance of a wire.

• Understanding the principle of a potentiometer and its use for

(i) comparing the emf’s of two primary cells.

(ii) finding the internal resistance of a cell.

• Understanding the principle of a moving coil galvanometer and knowing the meaning of its ‘figure of merit’.

• Using the ‘half deflection method’ for finding the resistance of a given moving coil galvanometer.

• Understanding the meaning of ‘parallex’; its use in finding the focal length of a (i) convex mirror (ii) concave mirror and (iii) convex lens.

• Learning how to plot, the angle of deviation versus the angle of incidence, for a given triangular prism.

• Understanding the use of a travelling microscope, for finding the ‘real’ and ‘apparent’ depth of a given glass slab; and hence the refractive index of the glass used.

• Understanding the meaning of forward and reverse biased pn junction.

• Studying the variation/ relation, between current flowing through a diode and the voltage applied across it.

• Studying I-V characterstics curve of a Zener diode.

• Understanding the role ofZener diode as a voltage regulator device.

• Learning to identify and characterize, diode, LED, transistor, IC, resistor, capacitor, from a mixed collection by:

(i) Physical examination

(ii) By making use of multimeter.

• Understanding the multiple functions of a multimeter.

• Learning the use of a multimeter for measuring / checking a variety of electronic components.

EXPERIMENT-1

Vernier Callipers: Its use to measure internal and external diameter and depth of a vessel.

CONCEPT MAP

Vernier Callipers

The essential parts of this device, invented by the French mathematician, Pierre Vernier, are shown in the figure below.

Leart Count of the Vernier

The use of an additional scale - called the vernier scale - enables this device to have a much better ’least count’ than the metre scale. The least count of a vernier callipers is given by the formula:

Least count $=(1$ main scale divistion - 1 vernier scale division $)$

$=\dfrac{(\text { value of the smallest division on the main scale })}{(\text { number of divisions on the vernier scale })}$

Reading the Vernier

The reading of the the vernier is a two step process:

(i) reading the main scale value just before the zero of the vernier scale and

(ii) finding the number (n) of the vernier scale division that just coincides with some main scale division.

The total reading equals

Main scale reading $+(\mathrm{n})$ (least count of vernier)

Thus in the fiture given here

The total reading of the vernier callipers (least count $=0.01 \mathrm{~cm}$ ) equals

$$ (1.9 \mathrm{~cm})+6 \times(0.01 \mathrm{~cm})=1.96 \mathrm{~cm} $$

‘Zero Error’

A given vernier callipers in said to have a ‘zero error’ if the zero of the vernier scale does not coincide with the zero of the main scale when the jaws of the vernier (just) touch each other.

‘Zero error’ can be positive or negative. It is positive when the zero of the vernier lies to the right of the zero of main scale. (when the jaws just touch other).

It is negative when the ‘zero’ of the vernier lies to the left of the zero of the main scale (when the jaws just touch each other). Zero error must always be algebrically subtracted from the observed reading.

It is important to remember that before using a vernier callipers, we

(i) Find / know the value of its one main scale division as well as its ’least count’

(ii) Check for the presence of any ‘zero error’ in it. If a zero error is present, we need to know both its magnitude and sign.

(iii) Always apply the relevant ‘zero correction’ to the readings taken by it.

Using the Vernier

For doing any experiment with the vernier, we first find

(i) its least count and

(ii) zero error, if any, in it.

These have to be used, in the manner already explained, in all readings taken with the vernier.

1. To measure the external diameter of a given spherical object, say, a given cylindrical vessel / glass marble.

After finding the least count and zero error, the given object is held between the ‘jaws’ of the vernier in the manner shown.

The reading are taken by holding the given object first in one-direction and then in a direction perpendicular to that. These ‘pair of readings’ are repeated at ‘at least three points’ on the object. The ‘zero-correction’, if needed, is applied to the mean of all the redings taken. This mean corrected value (appropriately ‘rounded off’, as per the least count of the given vernier) gives the diameter of the given object.

2. To measure the internal diameter and depth of a given (cylindrical) beaker / calorimeter.

We make use of the upper / outer jaws of the vernier callipers and the sliding strips provided at its back, for taking these readings. The relevant settings are shwon in the figures given below.

The readings, for the internal diamtere, are taken in the same way as those for the ’external diameter’. We again take readings in two mutually perpendicular directions at ‘at least three points’ in the beaker / calorimeter.

The readings for the ‘depth’ may be taken, only once each, at ‘at least three ponits’ within the beaker calorimeter. It is important here to keep the vernier in a vertical position and to keep the edge of the main scale on the mouth of the beaker / calorimeter. The zero correcion, if needed, has to be applied to the mean of the readings taken and the final result ‘rounded off’ as per the least count of the given vernier.

EXPERIMENT-2

Screw Gauge: Its use to determine thickness / diameter of thin sheet / wire.

Screw Gauge

A screw gauge - also sometimes known as the micrometer - has an accurately threaded screw having a closely fitting nut. The threadings, of the nut and the screw, match each other. When rotated, the screw not only rotates along a circular path but also advances along a linear path.

The essential parts, of the screw gauge, are shown, and labelled, in the figure given here.

Pitch

The pitch, of a screw gauge, equals the linear distance moved by the screw when it is given one complete rotation. It equals the separation between the successive threads of the screw.

Least Count

The minimum distance, that can be measured by a screw guage, is known as its least count.

The circular scale, in a screw gauge, has (usually) 100 divisions on it. The least count of a screw gauge is therefore, given by

$$ \text { Least count }=\dfrac{\text { Pitch }}{(\text { Number of divisons on the circular scale })} $$

The screw gauges, in common use, have a least count of $0.001 \mathrm{~cm}$. However, the more precise versions often have least counts of $0.0005 \mathrm{~cm}$.

Reading the Screw Guage

The (thin) sheet / wire, whose thickness / diameter is to be measured, is ‘just held’ between the faces of the (fixed) and A and the (movable) and B of the screw gauge. At this position, the reading of the main scale (say m), just before the end of its circular scale, is read. One also notes the number (n) of the circular scale division that just coincides with the line of graduation of its main scale. The total reading, of the screw gauge, is then

Totao reading $=(\mathrm{m}+\mathrm{n} \times$ least count $)$

Zero Error

Due to wear and tear, or due to some manufacturing defect, the screw gauge can have a zero error.

When the faces of the ends A and B of a screw gauge are put just in contact with each other, and the zero of the circular scale lies along the line of graduation of the main scale, the screw aguge does not have any zero error. This is shown in the figure below.

Quite often, however, the perfect, or ideal condition shown above, does not hold. The ‘zero’, of the circular scale, may lie below / above the ‘reference line’ (the line of graduation of the main scale) when the faces, of the ends A and B, are just in contact. When the ‘zero’, lies below the reference line, the screw gauge is said to have a positive zero error. Whe, the ‘zero’ lies above the reference line, the screw gauge is said to have a negative zero error.

The magnitude, of the zero error, equals the product of the least count (of the screw gauge) with the number (with reference to its zero) of circular scale division that coincides with the reference line. Thus in the figures above, the magnitudes, of the zero error, are $(6 \times$ L.C $)$ and $(100-92$, i.e. 8$) \times$ L.C.

The zero error is always subtracted algebraically from the observed readings.

Backlash Error

The ‘wear and tear’, present in an ‘in-use’ screw gauge, can bring in some ‘play’ in the motion of its screw. In such cases, the screw may not move forward even when the screw head is being rotated. This error is known as ‘backlash error’.

It can be avoided / minimized by always rotating the screw in one direction only.

EXPERIMENT-3

Simple Pendulum: Dissipation of energy by plotting a graph between square of amplitude and time.

CONCEPT MAP

Simple Penculum

An ‘ideal’ simple pendulum has a ‘point’ mass, suspended from a weightless, inextensible string.

In practice, it is a good quality sewing thread that usually serves the purpose of a ‘weightless’, inextensible string. The ‘point mass’ is usually a small heavy metal sphere (known as the bob) that has a hook for attaching it to the thread. The other end of this thread is put in the cut of a split cork; this cork is firmly held in a suitable stand.

We take the distance between the point of suspension, and the centre of mass of the bobo, as the ’length’ of a given simple pendulum.

Nature of Motion of a Simple Pendulum

The bob, of a simple pendulum, oscillates in a ‘periodic way’ when it is displaced to one side (from its normal equilibrium position) and ’let-go’. These periodic oscillations are (very nearly) simple harmonic oscillatinos only if the initial displacement (the amplitude) of the bob (from its equilibrium position) is quite small in comparision to the length of the simple pendulum. This can be seen as follows:

The equation of motion, of the angular oscillations, of the bob, is

$$ \mathrm{I} \dfrac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-(\mathrm{mg} \sin \theta) \ell $$

or $\quad\left(\mathrm{m} \ell^{2}\right) \dfrac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-(\mathrm{mg} \sin \theta) \ell$

or $\quad \dfrac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-\dfrac{\mathrm{g}}{\ell} \sin \theta$

$$ =-\omega^{2} \sin \theta $$

where $\omega^{2}=\dfrac{g}{\ell}$

If $\theta$ is small, we can take $\sin \theta \simeq \theta$. We then have

$$ \dfrac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}} \simeq-\omega^{2} \theta $$

This is the equation of a simple harmonic motion of time period, $T$, where

$$ \mathrm{T}=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{\ell}{\mathrm{g}}} $$

Energy of a Particle Executing SHM

The displacement, of a particle, executing SHM, is a sinosoidal function of time. If can have the form

$$ y=a \sin \omega t \quad(a=\text { amplitude }) $$

The velocity, and accleration, are, therefore, given by

$$ \mathrm{v}=\dfrac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \omega \cos \omega \mathrm{t}=\omega \sqrt{\mathrm{a}^{2}-\mathrm{y}^{2}} $$

and $\quad \mathrm{f}=\dfrac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{a} \omega^{2} \sin \omega \mathrm{t}=-\omega^{2} \mathrm{y}$

The oscillating particle has both K.E. and P.E., at different points, on its oscillation path. These are given by

K.E. $=\dfrac{1}{2} \mathrm{mv}^{2}=\dfrac{1}{2} \mathrm{ma}^{2} \omega^{2} \cos ^{2} \omega \mathrm{t}$

$$ =\dfrac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right) $$

and P.E. $=\dfrac{1}{2} \mathrm{ma}^{2} \omega^{2} \sin ^{2} \omega \mathrm{t}$

$$ =\dfrac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2} $$

The total energy, having a constant value at all points and at all times, therefore, equals

$$ \mathrm{E}=\dfrac{1}{2} \mathrm{~m} \omega^{2} \mathrm{a}^{2} $$

For a pendulum of a given length, the total energy is, thereore, proportional to the square of the amplitude of its oscillations. We may, therefore, take the square of the amplitude, of the oscillations of a given simple pendulum, as an ‘indicator’ of its ’total energy’.

Dissipation of Energy with Time

A practical simple pendulum is never an ideal one; also it (usually) oscillates in air which can be regarded as a (mild) viscous medium. There is, therefore, some dissipation of enregy, by a (freely) oscillating simple pendulum. We observe this through a gradual but progressive decrease in the amplitude of its oscillations. We express this by saying that the (small amplitude) oscillations of a (practical) simple pendulum, (oscillating in air) are damped harmonic oscillations.

Damped Harmonic Oscillations

THe (mistantaneous) damping force, acting on an oscillating particle, is usually taken as proportional to its instantaneous velocity. The equation of motion, of damped oscillations, is, therefore, usually written in the form:

$$ \dfrac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}+2 \mathrm{~b} \dfrac{\mathrm{dy}}{\mathrm{dt}^{2}}+\mathrm{c}^{2} \mathrm{y}=0 $$

On solving this differential equation, we find that the

(i) Amplitude of oscillations decays exponentially with time:

$$ \mathrm{a}(\mathrm{t})=\mathrm{a}(0) \mathrm{e}^{-\alpha \mathrm{t}} $$

(ii) The time period of oscillations shows an increase; the exact magnitude of this increase depends on the magnitude of the damping force that comes in play.

The figures, given below, show the difference between undamped and damped oscillations.

Practical ‘Observation’ of the Variation of Amplitude of an Oscillating Simple Pendulum with Time

A simple method, of observing the variation of amplitude, with time, can be as follows. We draw a line on the floor, below the oscillating bob. We then mark, on this line, points (on both sides) at distances of $1 \mathrm{~cm}, 2$ $\mathrm{cm}, 3 \mathrm{~cm}, 4 \mathrm{~cm}$, and $5 \mathrm{~cm}$ from the mean, or equilibrium, position. Set the bob oscillating and determine its time period.

Start with an amplitude of $5 \mathrm{~cm}$. Count the number of oscillations the pendulum makes after which it amplitude becomes $4 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}, 1 \mathrm{~cm}$ and (nearly) zero. These number of oscillations, multiplied by the time period (assumed to remain constant, (which it (nearly) is) given the times taken for the amplitude to reduce from $5 \mathrm{~cm}$ to its lower values. One can then prepare a table, of the form shown below, and use it to draw a graph between ‘square of amplitude’ and time.

Time Period of the Simple Pendulum $=\left(t _{0}\right) s$

S. No.	Amplitude	No. of oscillations taken for the amplitude to have this value	Corresponding value of time (in s)
1	$5 \mathrm{~cm}$	0	0
2	$4 \mathrm{~cm}$	$\mathrm{n} _{1}$	$\mathrm{n} _{1}\left(\mathrm{t} _{\theta}\right)$
3	$3 \mathrm{~cm}$	$\mathrm{n} _{2}$	$\mathrm{n} _{2}\left(\mathrm{t} _{\theta}\right)$
4	$2 \mathrm{~cm}$	$\mathrm{n} _{3}$	$\mathrm{n} _{3}\left(\mathrm{t} _{\theta}\right)$
5	$1 \mathrm{~cm}$	$\mathrm{n} _{4}$	$\mathrm{n} _{4}\left(\mathrm{t} _{\theta}\right)$
6	(nearly) zero	$\mathrm{n} _{5}$	$\mathrm{n} _{5}\left(\mathrm{t} _{\theta}\right)$

Table for Graph

S. No.	1	2	3	4	5	6
Square of Amplitude	$25 \mathrm{~cm}^{2}$	$16 \mathrm{~cm}^{2}$	$9 \mathrm{~cm}^{2}$	$4 \mathrm{~cm}^{2}$	$1 \mathrm{~cm}^{2}$	$0 \mathrm{~cm}^{2}$
Time $\longrightarrow$	—	—	—	—	—	—

The graph is likely to show an exponential decay of square of amplitude ( $\alpha$ energy) with time.

EXPERIMENT-4

Meter Scale: Mass of a given object by principle of moments.

CONCEPT MAP

Moment of a Force

The effectiveness of a given force, in producing rotation, of a given object, about some fixed axis (or point), is measured through ’the moment of the given force’, in the given set up.

We define the magnitude of the moment of a force as the product of the magnitude of the force and the ‘force-arm’. Here ‘force-arm’ equals the perpendicular distance between the ’line of action’ of the force and the axis (or point) about which rotation is taking place.

Magnitude of the moment of force $=\mathrm{Fr}$

Moment, of a force, (about a given axis or point) is a vector quantity. Its direction is perpendicular to the plane defined by the vector $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{r}}$, and is in the sense of rotation of a right handed screw rotated from $\overrightarrow{\mathrm{r}}$ to $\overrightarrow{\mathrm{F}}$. We may, therefore, express the moment of a force (say $\vec{\tau}$ ) through the vector relation

$$ \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}} $$

Here $\overrightarrow{\mathrm{r}}$ is the position vector, of the point of application of the force $(\overrightarrow{\mathrm{F}})$; with respect to the point ( $\mathrm{O}$, taken as the origin) about which rotation is taking place.

For rotation about an axis, the origin is taken as the point, where the perpendicular, drawn from the point of application $(\mathrm{P})$ of the force, on the axis of rotation, meets this axis.

In general, then

$$ |\vec{\tau}|=\mathrm{rF} \sin \theta $$

Here $r$ and $F$ denote the magnitudes of the vector $\vec{r}$ and $\vec{F}$; also $\theta$ is the angle between vectors $\vec{r}$ and $\vec{F}$.

Principle of Moments

We state the principle (or law) of moments as follows:

Whenever a body is in equilibrium, under the action of a number of coplanar forces, the sum of the anticlockwise moments, about any arbitrary point, is equal to the sum of the clockwise moments about the same point.

Alternatively, if a body is in equilibrium under the action of a numbe rof coplanar forces, the algebraic sum of all the moments, about any arbitrary point, is zero.

Practicle Application of the Principle of Moments

The principle of moments can be used to find the mass of a given object by using a meter scale, a sharp wedge and suitable known weights. With a slight modification, this ‘set-up’ can be used to find the mass of the meter scale itself.

For finding the (unknown) mass, say $m _{x}$, of an object $x$, the ‘set-up’ can be arranged as follows:

We then have

$$ \begin{aligned} x a & =w b \\ \text { or } \quad x & =w\left(\dfrac{b}{a}\right) \end{aligned} $$

We may also use a set-up in which two known weights are used. This type of set up is useful when the unknown mass is a heavy mass.

In this case, we have

$$ \begin{aligned} & \mathrm{xa}=\mathrm{w} _{1} \mathrm{~b}+\mathrm{w} _{2} \mathrm{c} \\ & \therefore \quad \mathrm{x}=\left(\dfrac{\mathrm{w} _{1} \mathrm{~b}+\mathrm{w} _{2} \mathrm{c}}{\mathrm{a}}\right) \end{aligned} $$

This set-up can be used to ‘weigh’ the meter scale itself. In this case, the wedge is put at an ‘off-centre’ point and two known masses are put on the two sides of the wedge and adjusted to ‘balance the system’.

In this case, if $m$ is the mass of the meter scale, we have

$$ \begin{aligned} & \mathrm{m} _{1} \mathrm{a}=\mathrm{m} _{1}(50-\mathrm{a})+\mathrm{m} _{2} \mathrm{~b} \\ & \therefore \mathrm{m}=\dfrac{\mathrm{m} _{1} \mathrm{a}-\mathrm{m} _{2} \mathrm{~b}}{(50-\mathrm{a})} \end{aligned} $$

When the meter scale is (horizontally) balanced on the wedge (by adjusting) the position of the masses $x$ and $w$ ), we have

(Algebraic) sum total of all moments, about the wedge $=$ zero

$\therefore \mathrm{m} _{\mathrm{x}} \mathrm{ga}-\mathrm{m} _{\mathrm{w}} \mathrm{gb}=0$

or $\quad \mathrm{m} _{\mathrm{x}}=\dfrac{\mathrm{m} _{\mathrm{w}} \cdot \mathrm{b}}{\mathrm{a}}$

The unknown mass $\mathrm{m} _{\mathrm{x}}$ can this be easily calculated.

To use the second ‘set-up’ to find the mass $\left(m _{s}\right)$ of the metre scale itself, the wedge is shifted from its usual ‘central’ position. We now use known masses, $\mathrm{m} _{\mathrm{w}}$ and $\mathrm{m} _{\mathrm{w} 1}$ and $\mathrm{m} _{\mathrm{w} 2}$. Let $\mathrm{m} _{\mathrm{s}}$ be the mass of the meter scale itself. When the metre scale gets (horizontally) balanced, we again have

(Algebraic) sum total of moment of all the weights about the wedge $=0$

$$ \begin{aligned} & \therefore \quad m _{w} \times a-m _{s} \cdot d-m _{w 1} \times b-m _{w 2} \times c=0 \\ & \text { or } \quad m _{s} d=m _{w} \times a-m _{w 1} \times b-m _{w 2} c \\ & \text { or } \quad m _{s}=\left[\dfrac{m _{w} \times a-m _{w 1} b-m _{w 2} c}{d}\right] \end{aligned} $$

It may be noted that the two ‘weights’, $\mathrm{w} _{1}$ and $\mathrm{w} _{2}$, shown on the right, can be (easily) replaced by a suitable single weight.

EXPERIMENT-5

Young’s Modulus of Elasticity of the Material of a Wire.

CONCEPT MAP

Elasticity

Elasticity is the property of objects of “tending to regain their original conditions after the removal of the deforming forces”.

Stress

When an object is deformed from its normal (equilibrium) condition, internal forces get ‘set-up’ in it that tend to oppose that deformation. This internal recovering force, measured per unit area, is known as the stress developed in the (deformed) object.

The SI units of stress are, therefore, $\dfrac{\mathrm{N}}{\mathrm{m}^{2}}$. Its dimensions are $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$.

Strain

When an object is subjected to the action of an external deforming force, the change in the dimensions concerned, measured per unit value of that dimension, is called strain. Strain is a dimensionless quantity; it therefore, has no units associated with it.

Hook’s Law

The fundamental law of elasticity, formulated by Robert Hook, is known as Hook’s law. According to this law:

“For small values of strain, the stress developed in an object, is proportional to the strain produced in it”. The limit, up to which this proportionality between stress and strain holds for a given material, is known as the ’elastic limit’ of that material.

Elastic Module

Within the elastic limits, the stress developed in a material, is proportional to the strain produced in it. Thus, within the elastic limit,

$$ \dfrac{\text { Stress }}{\text { Strain }}=a \text { constant } $$

This constant is a characterstic of the material involved and is known as an ’elastic modulus’ of that material.

There are three types of elastic moduli that are defined for a given material. When a material is subjected to a longitudinal strain, the ratio, of the longitudinal stress to the longitudinal strain, is known as the Young’s modulus $(\mathrm{Y})$ for that material. Thus

$$ \text { Young’s modulus }=Y=\dfrac{\text { Longitudinal stress }}{\text { Longitudinal strain }} $$

In case of a volume strain, the ratio of the volume stress to the volume strain is known as the ‘Bulk Modulus’ of the material. Thus

$$ \text { Bulk Modulus }(\mathrm{K})=\dfrac{\text { Volume stress }}{\text { Volumestrain }} $$

Sometimes a material is so strained so as to cause a change in its shape without any change in its volume. We then speak of the modulus of rigidity $(\eta)$ of the material.

Modulus of rigidity $=\eta=\dfrac{\text { Shearing stress }}{\text { Shear strain }}$

The SI units of all these moduli of elasticity is $\dfrac{\mathrm{N}}{\mathrm{m}^{2}}$.

In addition to these three moduli of elasticity, we also define a dimensionless ratio, called Poisson’s ratio $(\sigma)$ for a given material. Poisson’s ratio is the ratio of the lateral strain to the longitudinal strain. For a wire, lateral strain equals $\left(\dfrac{\Delta \mathrm{D}}{\mathrm{D}}\right)$ where $\mathrm{D}$ is the diameter of the wire. $\sigma=\dfrac{-(\Delta \mathrm{D} / \mathrm{D})}{(\Delta \mathrm{L} / \mathrm{L})}$

The four elastic ‘constants’, $Y, K, \eta$ and $\sigma$ have two relations connecting them. We thus effectively have only two independent elastic constant for a material.

Searle’s Apparatus

Searle’s apparatus is a convenient set up that is often used in laboratory for determining Young’s modulus of the material of a given wire.

A labelled diagram of Searle’s apparatus is given here.

The arrangement, used in this apparatus, to find the (small) change in length of the experimental wire. It has a spirit level, supported horizontally. One end of this spirit level rests on an arm that is fixed to one of the metal frames. Its other end rests on the tip of the screw of a spherometer set-up. The readings of this spherometer, corresponding to the ‘in-centre’ positions of the bubble of the sprit level, without and with the load, on the experimental wire, enable us to find the ‘change in length’ of the experimental wire. These readings, along with readings for the diameter of the wire, enable us to calculate the Young’s modulus for the material of the wire.

Relevant Formula

Let a load $\mathrm{W}$ be applied on the experimental wire. Let $\mathrm{d}$ be the diameter of this wire and $L$ its unstretched (original) length. If $\Delta \mathrm{L}$ be the change in length, due to the load W, we have

Longitudinal stress $=\dfrac{\mathrm{W}}{\pi\left(\dfrac{\mathrm{d}}{2}\right)^{2}}=\dfrac{4 \mathrm{~W}}{\pi \mathrm{d}^{2}}$

and longitudinal strain $=\left(\dfrac{\Delta \mathrm{L}}{\mathrm{L}}\right)$

$\therefore \quad \mathrm{Y}=\dfrac{\text { Longitudinal stress }}{\text { Longitudinal strain }}$

$$ \begin{aligned} & =\dfrac{4 \mathrm{~W}}{\pi \mathrm{d}^{2}} \times \dfrac{1}{\left(\dfrac{\Delta \mathrm{L}}{\mathrm{L}}\right)} \\ & =\dfrac{4 \mathrm{WL}}{\pi \mathrm{d}^{2}(\Delta \mathrm{L})} \end{aligned} $$

This relation shows that a graph between $\mathrm{W}$ and $\Delta \mathrm{L}$ would be a straight line whose slope, can be used to calculate the mean value of $\mathrm{Y}$ corresponding to a given set of readings.

EXPERIMENT-6

Surface tension of water by capillary rise and effect of detergents.

CONCEPT MAP

Inter Atomic / Inter Molecular Forces

These are the forces that the atoms/molecules, of a substance, exert on one another. They are short range attractive forces; however, they become repulsive if the atoms / molecules are too close to each other.

Cohesive Forces

Forces, between atoms / molecules, of the same nature, are known as cohesive forces.

Adhesive Forces

Forces, between atoms /molecules of different nature, are known as adhesive forces.

Surface Tension

The surface of a liquid behaves as if it is in a state of stress. The properties, associated with a liquid surface, indicate that each side of the liquid, on either side of an imaginary line on the liquid surface, appears to be pulling the other side towards itself. We call this phenomenon as the phenonmenon of surface tension.

Quantitative Definition of ‘Surface Tension’

Imagine a line, say $A B$, to be drawn on the surface of a liquid. The (pulling) force, per unit length of this line, directed perpendicular to this line, and in the plane of the (liquid) surface, is taken as a measure of the surface tension of the given liquid.

The Si units of surface tension are N/m (newton per meter). Its dimensions are $\left[\mathrm{MT}^{-2}\right]$.

$\mathrm{F} _{1}=$ Pulling force, on side I, due to side II.

$\mathrm{F} _{2}=$ Pulling force, on side II, due to side I.

Angle of Contact

The surface of a liquid, when put in a container, acquires a curved shape. The extent and nature, of this ‘curving’, depends on the liquid and the material of the container.

We use a quantity, called ‘angle of contact’, as a measure of the ‘curved meniscus’ of a liquid in a given container. The angle of contact is the angle between the tangents to the liquid surface and the surface of the container. For a concave meniscus (like that of water in a glass tube), this angle is taken as the relevant acute angle while for a convex meniscus (like that of mercury in a glass tube), it is taken as the relevant obtuse angle.

Surface Energy

The potential energy, of the molecules on the surface of a liquid, is more than that of the molecules in its interior. This (extra) potential energy, per unit area of the liquid surface, is known as the surface energy per unit area of the given liquid.

The units of surface energy per unit area, are joule / meter squared $\left(\mathrm{J} / \mathrm{m}^{2}\right)$ while its dimensions are $\left[\mathrm{MT}^{-2}\right]$.

It is easty to see that the surface energy, per unit area of a given liquid, equals the ‘surface tension’ of that liquid.

Pressure Difference Across the Curved Meniscus of a Liquid

The pressure, on the concave side, of the curved meniscus of a liquid, is more thatn that on its other side. It can be shown that this excess pressure, (say $\Delta \mathrm{P}$ ), is given by

$$ \Delta \mathrm{P}=\dfrac{2 \mathrm{~T}}{\mathrm{r}} $$

Here $\mathrm{T}=$ surface tension of the given liquid and $\mathrm{r}=$ radius of curvature of its curved meniscus.

For a bubble, blown out of a given liquid (like say, saop solution), the excess pressure is double that for a liquid drop. This is because, the boundary surface (between) the liquid and air) exists both on the inner as well as the outer side of the bubble.

Thus, excess pressure, inside a bubble, equals $\dfrac{4 \mathrm{~T}}{\mathrm{r}}$.

Capillary Tubes

Tubes, of very small diameter (of the order of a mm or so), are known as capillary tubes.

‘Rise’ of Liquid in a Capillary Tube

When a capillary tube is dipped and held vertically, in a container of a given liquid, the level of the liquid, in the capillary tube, is different from its level in the outiside container. This difference, in liquid levels, say h, is related to the surface tension $(T)$, and the density $(\rho)$, of the liquid, through the relation.

$$ \mathrm{T}=\dfrac{\mathrm{r}\left(\mathrm{h}+\dfrac{1}{3} \mathrm{r}\right) \rho \mathrm{g}}{2 \cos \theta} $$

Here $\mathrm{r}=$ radius of the capillary tube;

$\mathrm{g}=$ acceleration due to gravity

and $\theta=$ angle of contact between the liquid and the walls of the container.

We can take $\theta=0^{0}$ when the liquid gets ‘spread out’ on the walls of the container. For such cases (e.g. water in a glass tube), we have $\cos \theta \simeq 1$. Hence the above relation takes the form

$$ \mathrm{T}=\dfrac{\mathrm{r}\left(\mathrm{h}+\dfrac{1}{3} \mathrm{r}\right) \rho \mathrm{g}}{2} $$

Quite often $\mathrm{r}«$ h. We can then write

$$ \mathrm{T} \simeq \dfrac{\mathrm{rh} \rho \mathrm{g}}{2} $$

It is this relation that we generally use for finding the surface tension of a liquid by the capillary rise method.

Travelling Microscope

We generally use a ‘special microscope’, called a travelling microscope, for finding the surface tension of a liquid by the capillary rise method. The ‘stands’, attached to the tube of the ‘compound microscope’used here are such taht the tube can be moved both ‘up and down’ (i.e. vertically) as well as ‘right and left’ (i.e., horizontally). There are ‘scales’ (‘main scale’, with its attached vernier scale) that enable one to measure (the horizontal or vertical) movement of the ‘cross wires’ in the ’eye-piece’ of this microscope. The ’least count’ of the vernier, used here, is generally $0.0005 \mathrm{~cm}$ or $0.001 \mathrm{~cm}$.

We use this microscope for measuring the

(i) ‘height’ of ‘rise’, of the liquid in the capillary tube.

(ii) the diameter of the capillary tube.

It is important here to note that

(i) liquids, like water (in a glass tube), ‘rise’ above the ‘outside level’, in a capillary tube.

(ii) liquids, like mercury (in a glass tube), ‘fall’ below the ‘outside level’, in a capillary tube.

Thus ’ $h$ ’ is ‘positive’, for liquids, like water and ’negative’, for liquids, like mercury (in a glass capillary tube).

‘Experimental Set-up’

A diagram, of the experimental set-up used in the exeperiment for ‘Measuring the surface tension of water, by the capillary rise’ method is given below.

We use this ‘set-up’ for measuing ‘h’ as well as ’ $\mathrm{r}$ ‘. The angle of contact, for water is a glass capillary tube, is generally taken as $0^{0}$. Using the value of $\rho$ (density of water), and $g$ (acceleration due to gravity), from standard tables, one can calculate the surface tension of water (say, $\mathrm{T}$ ) through the relation

$$ \mathrm{T}=\dfrac{\mathrm{r}\left(\mathrm{h}+\dfrac{1}{3} \mathrm{r}\right) \rho \mathrm{g}}{2} $$

Effect of Detergents

The addition of a deteregent to water, leads to a decrease is the surface tension of water.

Detergents help in cleaning of dirty or greasy clothes. These action can be understand in two ways:

(i) It is energatically favourable to have globs of dirt surounded by detergent and then water.

(ii) The lowering of surface tension, because of the addition of detergents, leads to a better ‘capillary rise’ through the pores in the clothes. This in turn, leads to a better cleaning of the clothes.

EXPERIMENT-7

Coefficient of viscosity, of a given viscous liquid, by measuring terminal velocity of a given spherical body.

CONCEPT MAP

Viscosity

There exists some ‘internal friction’, between adjacent fluid layers, when the fluid is in motion. This ‘internal friction’, which tends to oppose any relative motion between adjacent fluid layers, is called viscosity.

Newton’s Law of Viscous Flow

Let there be two adjacent fluid layers, that are at distances $z$ and $(z+d z)$, from the reference stationary layer. Let the speed of the flow, in these two layers be $v$ and $(v+d v)$. If A is the (in contact) surface area of these two fluid layers, and $\mathrm{F}$ the retarding viscous force, between these two layers, it turns out

$\mathrm{F} \propto \mathrm{A}$

and $\quad \mathrm{F} \propto \dfrac{\mathrm{dv}}{\mathrm{dz}}$

We can, therefore, write

$$ \mathrm{F}=-\eta \mathrm{A} \dfrac{\mathrm{dv}}{\mathrm{dz}} $$

We refer to this result as Newton’s law of viscosity.

Here $\eta$ is a constant of proportionality, that is characterstics of the fluid considered. We call $\eta$ as the coefficient of viscosity of the given liquid.

Coefficient of Viscosity

From Newton’s law, it follows that

$$ |F|=\eta $$

If $\mathrm{A}=1$ unit and $\dfrac{\mathrm{dv}}{\mathrm{dz}}=1$ unit

We may, therefore, say;

The coefficient of viscosity, of a given fluid, is the (magnitude of) force required to maintain a unit velocity gradient between adjacent layers of unit contact area.

The dimensions of $\eta$ are

$$ [\eta]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] $$

The cgs unit of $\eta$ is known as the poise while its SI unit has been named as Pa-s (pascal-second). We have

poise $=\mathrm{g} \mathrm{cm}^{-1} \mathrm{~s}^{-1}$ or dyne $\mathrm{cm}^{-2} \mathrm{~s}$

and $\rho \mathrm{a}-\mathrm{s}=\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-1}$ or $\mathrm{Nm}^{-2} \mathrm{~s}$ or pascal-second

It is easy to see that

$$ 1 \mathrm{~Pa}-\mathrm{s}=10 \text { poise } $$

Stokes’ Law

When an object moves through a viscous fluid, the force of viscosity, that tends to oppose its motion, increases with an increse in its velocity. It was Stokes who obtained a formula for this force of viscosity. The result, obtained by him, is known as Stokes’ law.

According to Stokes’ law:

$$ \mathrm{F}=6 \pi \eta \mathrm{rv} $$

Here $\mathrm{F}$ is the force of viscosity (or viscous force) that acts on an object of size $r$ (a spherical ball of radius r) moving with a velocity $v$ in a medium having a coefficient of viscosity equal to $\eta$. The medium is supposed to be infinite in extent.

Stokes’s law can be easily obtained using the method of ‘dimensional analysis’.

Terminal Velocity

Let a (small) object (say, a (small) spherical ball, of radius $r$ ) fall vertically through a viscous medium having $\eta$ as its coefficient of viscosity. The net gravitational force on it, at first, accelerates it and, thereby, increases its velocity. The viscous force, acting on it, keeps on increasing (as per Stokes’ law) with this increase in its velocity. At a certain value of its velocity ( $\left(\mathrm{say} _{\mathrm{T}}\right.$ ), the viscous force becomes just equal (and opposite) to the (net) gravitational force acting on the object. The overall net force, acting on the object, then becomes zero and the object does not get accelerated any further. The velocity of the object then becomes constant at this limiting value $\left(\mathrm{v} _{\mathrm{T}}\right)$ of its velocity. Any further ‘fall’ of the object is then a ‘fall’ with a constant velocity (equal to $\mathrm{v} _{\mathrm{T}}$ ). We, therefore, call $\mathrm{v} _{\mathrm{T}}$ as the ’terminal velocity’ of the falling object.

‘A Formula’ for ‘Terminal Velocity

Let a (small) spherical ball, of radius $r$, (made up from a material of density $\rho$ ), fall vertically through a viscous medium (of density $\sigma$, say), having a coefficient of viscosity, $\eta$. The various forces acting on this spherical ball are:

(i) Force of ‘gravity’; this equals the ‘weight’ of the object and is equal to $\left(\dfrac{4 \pi}{3} r^{3} \rho g\right)$

(ii) Bouyant force, this equals the weight of that volume of the medium that gets displaced by the object;

$$ \text { ; it is equal to }\left(\dfrac{4 \pi}{3} \mathrm{r}^{3} \sigma \mathrm{g}\right) $$

(iii) Viscous force; as per Stokes’ law this equals $(6 \pi \eta \mathrm{rv})$ when the velocity of the object equals $\mathrm{v}$.

The net (downward) force, acting on the object, is

$\therefore F _{d}=\left(\dfrac{4 \pi}{3} r^{3} \rho g\right)-\left[\left(\dfrac{4 \pi}{3} r^{3} \sigma g\right)+(6 \pi \eta r v)\right]$

For $\mathrm{v}=\mathrm{v} _{\mathrm{T}}$, this net (downward) force becomes zero. Hence

$$ \begin{aligned} & \dfrac{4 \pi}{3} r^{3}(\rho-\sigma) g-6 \pi \eta r \mathrm{v} _{\mathrm{T}}=0 \\ \therefore \quad \mathrm{v} _{\mathrm{T}} & =\dfrac{4 \pi}{3} \dfrac{\mathrm{r}^{3}(\rho-\sigma) g}{(6 \pi \eta r)} \\ & =\left[\dfrac{2}{9 \eta}(\rho-\sigma) g\right] \mathrm{r}^{2} \end{aligned} $$

Thus $\mathrm{v} _{\mathrm{T}} \propto \mathrm{r}^{2}$

The slope (say, $\mathrm{m}$ ), of the graph between $\mathrm{v} _{\mathrm{T}}$ (on $\mathrm{y}$-axis) and $\mathrm{r}^{2}$ (on the $\mathrm{x}$-axis), is

$$ \begin{aligned} & m=\dfrac{2}{9 \eta}(\rho-\sigma) g \\ & \therefore \quad \eta=\dfrac{2}{9 m}(\rho-\sigma) g \end{aligned} $$

We can, therfore, calcualte $\eta$ by plotting a graph between $\mathrm{v} _{\mathrm{T}}$ and $\mathrm{r}^{2}$.

Experimental Set-up

The experimental set-up, used for finding that coefficient of visocisty of a viscous liquid (like glycerine), using (small) steel balls,of different radii, is shown below.

EXPERIMENT-8

Plotting a cooling curve between the temperature of a hot body and time.

CONCEPT MAP

Basic Heat Transfer Formula

The basic formula, for ‘heat transfer’, is

$$ \mathrm{H}=\mathrm{ms} \theta $$

Here $\mathrm{H}=$ heat lost / gained by an object of mass $\mathrm{m}$, made of a material of specific heat capacity, $\mathrm{s}$. The resulting fall/rise in temperature of the object, equals $\theta$.

Three Processes of Heat-Transfer

Heat can get transferred, from one object to another, via three processes. There are:

1. Conduction: This is the main method of heat transfer in solids.

2. Convection: This is generally the main method of heat transfer in fluids - liquids and gases.

3. Radiation: This is the process through which heat can get transferred even across vacuum.

Cooling of a Hot Body

When a hot body is kept in ‘isolation’, it cools down by losing heat mainly through convection and rediation. The air, with which it would still be in contact, has very poor conductivity, hence the heat lost by it, through conduction, is almost negligible.

Emission of Heat by Radiation

The law, associated with the ’emission’ of heat by radiation, is Stefan’s law. This law, strictly valid for a perfectly black body, states:

“The heat radiation emitted, per unit area per second, by a perfecly black body, is directly proportional to the fourth power of its absolute (kelvin) temperature”.

In mathematical form:

$$ \mathrm{E}=\sigma \mathrm{T}^{4} $$

Here $\mathrm{E}=$ Total energy radiated per unit area per second

$\mathrm{T}=$ Absolute temperature of the body

$\sigma=$ Aconstant of proportionality, known as Stefan’s constant.

It may be noted that Stefan’s law is not a ’law of colling’; it is a law that governs the emission of radiation by a black body.

For a ‘black body’, at a temperature $\mathrm{T}$ kelvin, kept in surroundings that are at a temperature $\mathrm{T} _{0}$ kelvin, the net ‘heat’, radiated per unit area per second, is

$$ \Delta \mathrm{E}=\sigma \mathrm{T}^{4}-6 \mathrm{~T} _{0}^{4}=\sigma\left(\mathrm{T}^{4}-\mathrm{T} _{0}^{4}\right) $$

If $\mathrm{T}$ were close to $\mathrm{T} _{0}$, we would have

$$ \begin{aligned} \Delta \mathrm{E}=\sigma & \left(\mathrm{T}^{2}-\mathrm{T} _{0}^{2}\right)\left(\mathrm{T}^{2}+\mathrm{T} _{0}^{2}\right) \\ & =\sigma\left(\mathrm{T}-\mathrm{T} _{0}\right)\left(\mathrm{T}+\mathrm{T} _{0}\right)\left(\mathrm{T}^{2}+\mathrm{T} _{0}^{2}\right) \end{aligned} $$

$$ \begin{aligned} & \approx \sigma\left(\mathrm{T}-\mathrm{T} _{0}\right)\left(2 \mathrm{~T} _{0}\right)\left(2 \mathrm{~T} _{0}^{2}\right) \quad\left(\because\left(\mathrm{T}-\mathrm{T} _{0}\right)=\Delta \mathrm{T}, \text { a small quantity }\right) \\ & =\left(4 \mathrm{~T} _{0}^{3} \sigma\right)\left(\mathrm{T}-\mathrm{T} _{0}\right) \\ & \Delta \mathrm{E} \propto\left(\mathrm{T}-\mathrm{T} _{0}\right) \end{aligned} $$

Thus, under this approximation $\left(\mathrm{T}-\mathrm{T} _{0}=\Delta \mathrm{T}\right.$, a small quantity), the net heat radiated pe unit area, per second, is directly proportional to the difference in temperture, between the hot body and its surroundings. This approximated form of Stefan’s law is similar to the law of cooling given by Newton. However, it is not quite correct to regard Newton’s law (of cooling) as an approximation of Stefan’s law corresponding to small differences of temperature between the hot body and its surroundings.

Newton’s Law of Cooling

According to Newton’s law of cooling: “The rate of fall in temperature, of a hot body, is directly proportional to the difference in temperature between the hot body and its surroundings”.

The mathematical form of this law is

$$ \begin{aligned} -\dfrac{\mathrm{d} \theta}{\mathrm{dt}} & \propto\left(\theta-\theta _{0}\right) \\ \text { or } \quad & -\dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta _{0}\right) \end{aligned} $$

This mathematical form leads to the result

$$ \int \dfrac{-\mathrm{d} \theta}{\theta-\theta _{0}}=\mathrm{k} \int \mathrm{dt} $$

or

$$-\left[\ell \mathrm{n}\left(\theta-\theta_0\right)\right]=\mathrm{kt}+\mathrm{c}$$

Thus the time, $\Delta \mathrm{t}$, taken by the hot body to have its temperature fall, from $\theta _{1}$ to $\theta _{2}$, say, is given by

$$ \begin{aligned} & \ell \mathrm{n} \left(\theta _{2}-\theta _{0}\right)-\ell \mathrm{n} \left(\theta _{1}-\theta _{0}\right)=\Delta \mathrm{t} \\ & \text { or } \Delta \mathrm{t}=\ell \mathrm{n} \left[\dfrac{\left(\theta _{2}-\theta _{0}\right)}{\left(\theta _{1}-\theta _{0}\right)}\right] \\ & \text { or } \quad \Delta \mathrm{t}=2.303 \log \left[\dfrac{\left(\theta _{2}-\theta _{0}\right)}{\left(\theta _{1}-\theta _{0}\right)}\right] \end{aligned} $$

The result

$$ \left[\ell \mathrm{n}\left(\theta-\theta _{0}\right)\right]=\mathrm{kt}+\mathrm{c}^{\prime} $$

or

$$ \left[\log \left(\theta-\theta _{0}\right)\right]=-\mathrm{kt}+\mathrm{c}^{\prime} $$

tells us that the graph between $\log \left(\theta-\theta _{0}\right)$ and time should be a straight line.

If the experimental data leads to such a straight line graph, it may be regarded as verifying Newton’s law of cooling.

It is often said that Newton’s law of cooling holds only if the difference in temperature, between the hot body and its surroundings, is small. This ‘conclusion’ appears to come from the ‘fact’ that Stefan’s law approaches Newton’s law when the hot body temperature is only slightly above the temperature of its surroundings.

However, the above ‘conclusion’ is not quite correct. This is because, ‘hot bodies’, when kept in air, lose heat mainly through the process of convection. If we aid this process (of loss of heat by convection) by keeping the hot body near an open window, or under a fan, the validity of Newton’s law holds even for quite large difference in temperature between the hot body and its surroundings. Therefore, when we provide such ‘conditions’ in our experiment, the ‘cooling curve’ (temperature versus time graph) has the form shown here and the graph of $\left[\log \left(\theta-\theta _{0}\right)\right]$ versus [time] has the straight line form given above.

Experiment Set-up

The experimental set-up, for plotting the ‘cooling curve’, may use a calorimeter (without its lid), provided with a stirrer. A thermometer may be suspended in the calorimeter which may be kept near an open window or under a fan. Hot water, pourded into the calorimeter, needs to be constantly stirred and its temperature noted at (appropriate) regular intervals of time. The (temperature) versus (time) graph may then be plotted to get the ‘cooling curve’. One may also plot the $\left[\log \left(\theta-\theta _{0}\right)\right]$ versus (time) graph to check the validity, or otherwise, of Newton’s law of cooling.

EXPERIMENT-9

Speed of sound, in air, at room temperature, using a resonance tube.

CONCEPT MAP

Superposition of Waves

The phenomenon of mixing of two, or more, waves, with one another, to produce a ‘resultant’ wave, is known as the phenomenon of ‘superposition of waves’.

Stationary Wave

The name ‘stationary wave’ is given to the resultant wave that gets formed when two identical waves, propagating in opposite directions, get ‘superposed’ on each other.

The ‘form’ of a stationary wave does not appear to change with time. Such waves are characterised by the presence of

(i) Nodal points (’ $\mathrm{N}$ ’ points or nodes) that are points where the ‘medium’ is at ‘rest’, at all time instants.

(ii) Antinodal points (‘A’ points or antinodes), that are points of maximum disturbance at any given instant oftime.

In a stationary wave, there is always a nodal point between two antinodal points and vice-versa.

Any given antinodal (or nodal) point is separated from its immediate next antinodal (or nodal) point by a distance $\dfrac{\lambda}{2}$. The distance between an antinodal (nodal) point and its immediate next nodal (antinodal) point is $\dfrac{\lambda}{4}$.

Air Columns

We speak of an air column when we have a part of air, confined to definite dimensions, inside a hollow (usually cylindrical) tube.

We refer to such a tube, open at both ends, as an open tube. However, if the tube is closed at one end, it is known as a closed tube.

Stationary Waves in an ‘Open Tube’

We can get stationary waves in the air column inside an open tube. For this, we need a ‘source’ of waves. When such a source is kept at one end of an open tube, the ‘direct’ waves (sent by the source), and the waves reflected at the other open end, superpose inside the tube and form ‘stationary waves’ in the ‘contained air-column’.

Permissible Modes of Vibration of the Air-column inside an open tube.

The ‘stationary waves’, formed in the air-column in an open tube, have to satisfy the following requirements.

1. The two free ends (or open ends), of the tube must be antinodal points (or antinodes) of the stationary waves formed in the air-column.

2. There must be a nodal point between two antinodal points and vice-versa.

The first three permissible modes of vibration of the air-column inside an open tube are shown below.

$$\begin{array}{lll} \text { Here } \dfrac{\lambda}{2}=\mathrm{L} & \text { Here } 2 \cdot \dfrac{\lambda}{2}=\mathrm{L} & \text { Here } 3 \cdot \dfrac{\lambda}{2}=\mathrm{L} \\ \text { or } \quad \lambda=2 \mathrm{~L} & \text { or } \quad \lambda=\mathrm{L} & \text { or } \quad \lambda=\dfrac{2}{3} \mathrm{~L} \\ v=\dfrac{\mathrm{V}}{2 \mathrm{~L}} & \therefore \quad v=2 \cdot \dfrac{\mathrm{V}}{2 \mathrm{~L}} & \\ \begin{array}{l} \text { (Fundamentalmode } \\ \text { or first harmonic) } \end{array} & \text { (Second harmonic) } & v=3\left(\dfrac{\mathrm{V}}{2 \mathrm{~L}}\right) \ \text { (Thirdharmonic) } \end{array}$$

We thus obseve that the permissible frequencies of vibration, inside an open tube, can be all integral multiples of the frequency of vibration of its simplest, or fundamental mode of vibration.

Permissible Modes of Vibration of the Air-column inside a closed tube.

The ‘stationary waves’, formed in the air-column of a ‘closed tube’, have to satisfy the following requirements.

1. The free (or open), of the tube, must be an antinodal point while the closed end must be a nodal point.

2. There must be a nodal point between two antinodal points and vice-versa. The first three permissible modes of vibration of the air-column inside a closed

tube, are shown below.

Here $\dfrac{\lambda}{4}=\mathrm{L}$ Here $\dfrac{3 \lambda}{2}=\mathrm{L}$ Here $\quad \dfrac{5 \lambda}{4}=\mathrm{L}$

or $\quad \lambda=4 \mathrm{~L}$ or $\quad \lambda=\dfrac{4 \mathrm{~L}}{3}$ or $\quad \lambda=\dfrac{4}{5} \mathrm{~L}$

$\therefore \quad v=\dfrac{\mathrm{V}}{4 \mathrm{~L}}$ (Fundamental mode or first harmonic) $\therefore \quad v=3 \cdot \dfrac{\mathrm{V}}{4 \mathrm{~L}}$ (Third harmonic) $\therefore \quad v=5 \dfrac{\mathrm{V}}{4 \mathrm{~L}}$ (Fifth harmonic)

We thus observe that the permissible frequencies of vibration, inside a closed tue, can only be ODD multiples of the simplest, or fundamental, mode of vibration.

The Tuning Fork

The tuning fork is a source of sound that gives out a single pure note of constant frequency. It is, therefore, a standard oscillator of constant frequency.

The shape of the tuning fork is shown below. When it is held from its handle, and one of the its prongs is struck lightly against a rubber pad, the prongs start vibrating transversly, both moving inwards, or outwards, simultaneously. The tuning fork then produces its characterstics note of a constant frequency. The positions of the nodal, and antinodal, points of a vibrating tuning fork, are as shown.

Resonance

The phenomenon of resonance takes place when the frequency of the applied periodic force (used to vibrate a given system) matches the natural frequency of the system itself.

The Resonance Column

The simple set up, used for studying the phenomenon of resonance in a closed tube, has been given the name ‘resonance column’. The length of the air-column, in this set up is adjustable. We adjust the length of this air-column so that its permissible fundamental frequency, or a permissible multiple of that frequency, matches the characterstics (known) frequency of the tuning fork, used to set the air-column into vibrations. The phenomenon of resonance then takes place and one gets a loud, or maximum, sound. A measurement of the length of the air-column, corresponding to which resonance takes place, enables one to calculate the sped of sound in air.

The ‘resonance column’ is available in two simple forms, shown below. However, it is the first form, in which the effective length of the air column is adjusted by raising or lowering of the reservoir, that is used more often.

The first and second positions of fresonance, of the air-column in a resonance-column, are shown here.

End-Correction

The reflection of sound does not take place exactly at the open end of the tube. It takes place a little distance (usually denoted by e) above the open end of the tube. This distance ’ $e$ ’ is called the ’endcorrection’. The ’end-correction’ is very nearly equal to $(0.3 \mathrm{D})$ where $\mathrm{D}$ is the diameter of the tube.

Experimental Set-up

The experimental set-up, used in this experiment, is shown below.

Calculation of the Speed of Sound

We take a tuning fork of known frequency (say $n$ ) and find the lengths of the air column corresponding to the first and second positions of resonance. If these lengths are $\ell _{1}$ and $\ell _{2}$, respectively, we have

$$ \ell _{1}+\mathrm{e}=\dfrac{\lambda}{4} $$

and $\quad \ell _{2}+\mathrm{e}=\dfrac{3 \lambda}{4}$

$\therefore \quad\left(\ell _{2}+\mathrm{e}\right)-\left(\ell _{1}+\mathrm{e}\right)=\left(\dfrac{3 \lambda}{4}-\dfrac{\lambda}{4}\right)=\dfrac{\lambda}{2}$

$\therefore \quad \lambda=2\left(\ell _{2}-\ell _{1}\right)$

Now $\quad \mathrm{V}=\mathrm{n} \lambda$

Hence speed of sound is given by

$$ \mathrm{V}=2 \mathrm{n}\left(\ell _{2}-\ell _{1}\right) $$

[Note: We can also calculate the end correction (e). We have from above, $\left(\ell _{2}+\mathrm{e}\right)-3\left(\ell _{1}+\mathrm{e}\right)=\dfrac{3 \lambda}{4}-3\left(\dfrac{\lambda}{4}\right)=0$

$\therefore \ell _{2}-3 \ell _{1}=2 \mathrm{e}$

or $\left.\mathrm{e}=\dfrac{\left(\ell _{2}-3 \ell _{1}\right)}{2}\right]$

Dependence of Speed of Sound on Temperature

The dependence of the speed of sound, (in a gas) on temperature, is as per the formula

$$ \dfrac{\mathrm{V} _{\mathrm{t} _{2}}}{\mathrm{~V} _{\mathrm{t} _{1}}}=\sqrt{\dfrac{\mathrm{t} _{2}+273}{\mathrm{t} _{1}+273}}=\sqrt{\dfrac{\mathrm{T} _{2}}{\mathrm{~T} _{1}}} $$

This speed of sound (in a gas) varies as the square root of the absolute temperature (of the gas).

EXPERIMENT–10

Specific heat capacity of a given (i) solid, (ii) liquid, by the method of mixtures.

CONCEPT MAP

Heat

We may think of heat as a form of energy that gets exchanged between two bodies because of a difference in temperature between them.

The SI unit of heat is, therefore, the same as that of energy, i.e., joule (J)

Quantity of Heat

The amount of heat, gained or lost by a substance, depends on its (i) mass (ii) nature and the (iii) change in its temperature $(\theta)$.

We, therefore, write the formula:

$$ \mathrm{H}=\mathrm{ms} \theta $$

Here $s$ is a ‘constant’, characterstics of the substance. It is known as its ‘specific heat capacity’.

‘Specific Heat Capacity’

The specific heat capacity of a substance is defined as the quantity of heat needed to bring a unit change in the temperature of a unit mass of the substance. The SI unit of specific heat capacity is $\mathrm{J} /(\mathrm{kg}-\mathrm{K})$. This also equals $\mathrm{J} /\left(\mathrm{kg}-{ }^{\circ} \mathrm{C}\right)$.

Principle of Calorimetery

Let two bodies, at different temperatures, be brought in contact with each other. The body, at a higher temperature, would then ’lose’ some heat which can be gained by the body at a lower temperature. This process of ‘heat exchange’ continues till the two bodies acquire the same common temperature.

The basic principle of ‘calorimetry’ is that

Heat gained by the ‘cold’ body = Heat lost by the ‘hot’ body.

This principle, therefore, implicity assumes that all the heat transferred remains confined within the two bodies. This is generally not quite true. However, in usual simple experiments on calorimetry, we assume this principle to be a valid principle.

Thermal Capacity or Heat Capacity

The thermal capacity, or heat capacity of a substance, is defined as the quantity of heat needed to change its temperature by a unit amount.

For a substance of mass $\mathrm{M}$

Thermal capacity $=\mathrm{M} \times$ its specific heat capacity

The SI unit of ’thermal capacity’ is $\mathrm{J} / \mathrm{K}$. It is also equal to $\mathrm{J} /{ }^{\circ} \mathrm{C}$.

Calorimeter

The ‘calorimeter’ is the usual simple instrument / device, used in laboratories for experiments on finding ‘specific heat capacity’ and other simple heat related quantities. The following diagram shows the details of construction for a ‘calorimeter’.

Method of Mixtures

In the ‘method of mixtures’ the ‘hot body’ is made to transfer its heat to the ‘cold body’, generally by using a calorimeter. This method is often used for finding the specific heat capacity of a solid as well as a liquid.

‘Specific Heat Capacity’ of a Solid

A calorimeter of known mass (say $m$ ), made from a material of known specific heat capacity (say $s$ ), is taken. A known mass of water (say $\mathrm{M} _{1}$ ) (or any other ‘suitable’ liquid of known specific heat capacity) is poured into it. The initial temperature (say $\theta _{0}$ ) of this water / liquid is noted. Let the speicfic heat capacity of water / liquid be $\mathrm{S} _{1}$.

A known mass of the solid (say $\mathrm{M} _{2}$ ), of unknown specific heat capacity (say $\mathrm{S} _{2}$ ), is heated to a temperature $\theta _{1}$. This solid is then put submerged into the water / liquid contained in the calorimeter. The ’liquid’ is continuously stirred. Let the final equalibrium temperature of the ’liquid’ be $\theta _{2}$.

We then have

Heat lost by the solid

$$ =\mathrm{M} _{2} \mathrm{~S} _{2}\left(\theta _{1}-\theta _{2}\right) $$

heat gained by the calorimeter and the water / liquid contained in it

$$ \begin{aligned} & =\operatorname{ms}\left(\theta _{2}-\theta _{0}\right)+\mathrm{M} _{1} \mathrm{~S} _{1}\left(\theta _{2}-\theta _{0}\right) \\ & =\left(\mathrm{ms}+\mathrm{M} _{1} \mathrm{~S} _{1}\right)\left(\theta _{2}-\theta _{0}\right) \end{aligned} $$

Putting heat lost $=$ Heat gained, we get

$$ \begin{aligned} & \mathrm{M} _{2} \mathrm{~S} _{2}\left(\theta _{1}-\theta _{2}\right)=\left(\mathrm{ms}+\mathrm{M} _{1} \mathrm{~S} _{1}\right)\left(\theta _{2}-\theta _{0}\right) \\ & \therefore \quad \mathrm{S} _{2}=\dfrac{\left(\mathrm{ms}+\mathrm{M} _{1} \mathrm{~S} _{1}\right)\left(\theta _{2}-\theta _{0}\right)}{\mathrm{M} _{2}\left(\theta _{1}-\theta _{2}\right)} \end{aligned} $$

We can thus calculate the (unknown) specific heat capacity of the solid.

[Note: The use of water / ‘suitable’ liquid, in the calorimeter, depends on the requirement that the given solid should neither dissolve in water / ‘suitable liquid’, nor react chemically with it].

Specific Heat Capacity of a Liquid

Here we can use a (suitable) solid made from a material of known specific heat capacity.

Let $\mathrm{M} _{1}=$ mass of solid taken

$\mathrm{S} _{1}=$ specific heat capacity of the material of the solid

$\theta _{1}=$ Initial temperature of the ‘heated up’ solid.

Also let $\mathrm{m}=$ mass of calorimeter

$\mathrm{s}=$ specific heat capacity of the material of the calorimeter

$\mathrm{M} _{2}=$ mass of liquid taken

$\mathrm{s} _{2}=$ (unknown) specific heat capacity of the liquid.

Also let $\theta _{0}$ be the initial temperature of the calorimeter and liquid; and $\theta _{2}=$ Final equilibrium temperature after the hot solid has been put / suspended into the liquid.

We now have

Heat lost by the solid $=\mathrm{M} _{1} \mathrm{~S} _{2}\left(\theta _{1}-\theta _{2}\right)$

Heat gained by calorimeter $=\operatorname{ms}\left(\theta _{2}-\theta _{0}\right)$

Heat gained by the liquid $=\mathrm{M} _{2} \mathrm{~S} _{2}\left(\theta _{2}-\theta _{0}\right)$

Putting,

Heat gained $=$ Heat lost, we get

$$ \begin{aligned} & \left(\mathrm{ms}+\mathrm{M} _{2} \mathrm{~S} _{2}\right)\left(\theta _{2}-\theta _{0}\right)=\mathrm{M} _{1} \mathrm{~S} _{1}\left(\theta _{1}-\theta _{2}\right) \\ \therefore \quad \mathrm{S} _{2}= & \dfrac{\left[\mathrm{M} _{1} \mathrm{~S} _{1}\left(\theta _{1}-\theta _{2}\right)-\mathrm{ms}\left(\theta _{2}-\theta _{0}\right)\right]}{\left[\mathrm{M} _{2}\left(\theta _{2}-\theta _{0}\right)\right]} \end{aligned} $$

We can use this formula to calculate the (unknown) specific heat capacity of the liquid.

[Note: We need to use such a ‘suitable’ solid that neither dissolves in the given liquid nor reacts chemically with it.]

EXPERIMENT-11

Metre Bridge : Determination of resistivity of the material of the given wire.

CONCEPT MAP

Metre Bridge

Aim: To determine the specific resistance, or resistivity, of the material of given wire, using a metre bridge.

Wheatstone’s Bridge

It is an arrangment of four resistance, say, P, Q, R and S (out of which P, Q and R are known and variable while $\mathrm{S}$ is normally unknown), which are connected to a source of electricity and a galvanometer, as shown in the diagram.

When the circuit is switched on, some current will flow through various branches of the circuit, including the galvanometer (of resistance $\mathrm{G}$ ).

Since P, Q and R are known and variable, we can adjust their values to make the current, and thereby the deflection, in the galvanometer reduced to zero.

This process of adjusting the values of $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$, to reduce the current through the galvanometer to zero (or to reduce the deflection in the galvanometer to zero), is called ‘balancing the bridge’. At balance $\mathrm{i} _{\mathrm{g}}=0$.

Condition for Balancing of Wheatstone’s Bridge

Applying Kirchoff’s voltage law for the loopABDA, we get:

$$ \begin{equation*} \mathrm{I} _{1}(\mathrm{P})+\mathrm{i} _{\mathrm{g}}(\mathrm{G})-\mathrm{I} _{2}(\mathrm{R})=0 \tag{1} \end{equation*} $$

Similarly, for loop BCDB, we get

$$ \begin{equation*} \left(\mathrm{I} _{1}-\mathrm{I} _{\mathrm{g}}\right) \mathrm{Q}-\left(\mathrm{I} _{2}+\mathrm{I} _{\mathrm{g}}\right) \mathrm{S}-\left(\mathrm{I} _{\mathrm{g}}\right) \mathrm{G}=0 \tag{2} \end{equation*} $$

When we make $\mathrm{I} _{\mathrm{g}}=0$, by adjusting $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$, equation (1) and equation (2) will become:

$$ \begin{equation*} \mathrm{I} _{1}(\mathrm{P})+0-\mathrm{I} _{2}(\mathrm{R})=0 \tag{3} \end{equation*} $$

or $\quad \mathrm{I} _{1}(\mathrm{P})=\mathrm{I} _{2}(\mathrm{R})$

and $\left(\mathrm{I} _{1}\right) \mathrm{Q}-\left(\mathrm{I} _{2}\right) \mathrm{S}-0=0$

or $\quad \mathrm{I}(\mathrm{Q})=\mathrm{I} _{2}(\mathrm{~S})$

Dividing (3) and (4), we have

$$ \begin{equation*} \dfrac{\mathrm{P}}{\mathrm{Q}}=\dfrac{\mathrm{R}}{\mathrm{S}} \tag{5} \end{equation*} $$

This equation $\dfrac{P}{Q}=\frac{R}{S}$ (at balance, i.e., when $\left(i _{g}=0\right)$ ) is called the Wheatstone Bridge’s principle or the condition for balancing of the Wheatstone’s bridge.

How to Find the value of $S$

From Wheatstone’s condition, we have

$\dfrac{\mathrm{P}}{\mathrm{Q}}=\dfrac{\mathrm{R}}{\mathrm{S}}$, at balance

$\therefore \mathrm{S}=\dfrac{\mathrm{R}}{\mathrm{P}} \mathrm{Q}$

where $\mathrm{S}$ is the unknown resistance and $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are known resistances.

Metre Bridge, a Modified Wheatstone’s Bridge

Wheatstone’s bridge consists of three known variable resistances, which are usually resisatnce boxes. Handling these individual reisistance boxes, along the unknown resistance wire, is not easy. So a modified bridges has been developed, which is easy to handle and convenient for doing experiments. This is called the metre bridge. The name ‘metre-bridge’ is derived from the use of a one ‘metre’ long uniform wire in this modified bridge.

Metre Bridge: The Apparatus

The metre bridge consists of a uniform resistance wire $\mathrm{AB}$ of length 1 metre ( $100 \mathrm{~cm}$ ), having a constant value of resistance per unit length (say, $\sigma \mathrm{ohm} \mathrm{cm}^{-1}$ ), throughout its length. This wire is fixed on a wooden board along with $\mathrm{L}-$ shaped metal strips at its two ends. There are terminals for connecting insulated copper wires, as shown in the diagram. A straight metal strip, fixed on the wooden board, with three terminals attached, provides two ‘gaps’. A metre scale is also attached, to meature lengths along the wire $\mathrm{AB}$, used for balancing the bridge.

Circuit Diagram of Metre Bridge

The given resistance wire, of unkonwn resistance ’ $\mathrm{X}$ ’ and length ’ $\mathrm{L}$ ‘, is connected in the gap I. A resistance box, which can have $\mathrm{R}=1 \Omega, 2 \Omega$, —— etc. is connected in gap 2 , as shown in the circuit diagram. A cell, of emf V, is connected between the terminals A and B, through a key.

From the terminal at $\mathrm{C}$, a galvanometer and a jockey $(\mathrm{J})$ are connected as shown. The jockey can be used to complete the bridge circuit by gently pressing it on to metre bridge wire, whenever it is required.

The above circuit of metre bridge is similar to the Wheatstone’s bridge circuit. The related Wheatstone’s bridge circuit is shown for better comparison.

The jockey, pressed on to the uniform wire $\mathrm{AB}$, divides the wire into two parts: (i) part $\mathrm{AJ}$ of resistance $(\ell \sigma)$ and (ii) part JB of resistance $(100-\ell) \sigma$.

Experimental Set-up and Procedure to Find $\mathbf{X}$

Using connecting wires, the resistance box, cell key, galvanometer and jockey are connected with the metre bridge as shown is circuit diagram. The given wire (of resistance $\mathrm{X}$ and length $\mathrm{L}$ ) is connected in the gap 1.

The correctness of the circuit is tested by presing the jockey, on to the uniform wire, first near terminal A and then near terminal B, after the cell key is switched on. If the deflections obtained in the galvanometer are in opposited directions, when the jockey is pressed at $\mathrm{A}$ and then at $\mathrm{B}$, the circuit made is correct and the experiment can be performed with this circuit.

After checking the correctness of the circuit, with R ohm in the resisatnce box, the jockey is pressed near terminal $\mathrm{A}$ and it is moved, gently, along the wire $\mathrm{AB}$, towards the terminal $\mathrm{B}$, in order to decrease the deflection in the galvanometer. The jockey is moved till the deflection in the galvanometer becomes zero. Here $\mathrm{i} _{\mathrm{g}}=0$, we say the bridge is balanced. The jockey position $(J)$ is called the balancing point.

The balancing length $\mathrm{AJ}=\ell$ is measured. The length $(100-\ell)$, is also obtained, in centimetre.

Determination of $\mathbf{X}$

According to condition for balancing of the bridge, we have

$$ \begin{align*} \dfrac{\mathrm{X}}{\mathrm{R}} & =\dfrac{\ell \sigma}{(100-\ell) \sigma} \\ \text { or } \quad \mathrm{X} & =\left(\dfrac{\ell}{100-\ell}\right) \mathrm{R} \tag{6} \end{align*} $$

As we know the value of R, (taken from the box), and measured value of $\ell$, we can calculate the value of unkonwn resistance $\mathrm{X}$ of the given wire, connected in gap $\mathrm{I}$.

A number of observations are to be taken corresponding to $\mathrm{R}=1 \Omega, 2 \Omega, 3 \Omega$, etc. each time calculating $\mathrm{X}$, using the corresponding balancing length in each case.

The mean of all the values of $X$, thus calculated, will give the most correct value of X.

Concept of Specific Resistance or Resistivity ( $\rho$ )

The resistance, say $X$, of a conductor, at constant temperature, depends on:

(i) the nature of its material

(ii) its length $(\ell)$ and

(iii) its area of cross section (A)

It is found that $\mathrm{X}$ varies directly as $\ell$ and $\mathrm{X}$ varies inversely and $\mathrm{A}$. Therefore, we have,

In equation (7), $\rho$ is called the specific resistance or resistivity; it depends on the nature of the material of the conductor and is a constant for a given material, at a given temperature. The unit of $\rho$ is ohm-metre.

If the conductor is in the form of a thin wire of average diameter (d), then

$$ \begin{equation*} \mathrm{A}=\frac{\pi \mathrm{d}^{2}}{4} \tag{8} \end{equation*} $$

From equations (7) and (8) we get

$$ \begin{equation*} \rho=\dfrac{\mathrm{X}}{\mathrm{L}}\left(\dfrac{\pi \mathrm{d}^{2}}{4}\right) \tag{9} \end{equation*} $$

Determination of ’ $d$ ’ using a Screw Gauge

The diameter (d) of the wire, of resistance (X) is measured using a screw guage, following the procedural steps explained in Experiment No. 2,

$$ \mathrm{d}=[\mathrm{M} . \mathrm{S} . \mathrm{R}+\text { (V.S.R) L.C }]-\text { Zero error } $$

(Note that $\mathrm{d}$, measured using screw guage, is in $\mathrm{mm}$.)

The length ‘L’ of the resistance wire can be measured with the help of a metre-scale.

Once the values of $X, d$ and $L$ are obtained, the specific resisance $(\rho)$, of the material of the resistance wire can be calculated using the equation

$$ \rho=\dfrac{X}{L}\left(\frac{\pi d^{2}}{4}\right) $$

The following points are considered in order to minizime the error in obtaining thevalue of X, using the metre-bridge.

1. The plug-keys of the resistance box are cleaned using sand paper.

2. All terminals and plug keys are made tight.

3. The circuit is not kept ‘on’ for a long time, as it may cause unnecessary heating.

4. The jockey is not dragged hard along the metre-bridge wire AB, as it may change the crosssectional area of the wire.

5. For better accuracy in the determination of $X$, the procedure can be repeated after putting the resistance wire in the right gap and the resistance box in the left gap. (The overall mean value of $X$ will give the most accurate result).

6. The value of R, put in the resistance box, may be adjusted in such a way that the balancing length $(\ell)$ is obtained around the $50 \mathrm{~cm}$ mark. The error, here, will be the least.

Example-1:

The following data was collected by a student to determine the resistance $\mathrm{X}$ of the given wire by metre bridge and hence to calculate the specific resistance ( $\rho$ ) of the material of the wire (along with the additional measurements for diameter and length of that wire). Find the value of $\rho$.

Sl. No.	Resistance in the right gap. $\mathrm{R}(\Omega)$	Balancing length measured from the left gap. $\ell(\mathrm{cm})$
1.	1	85.7
2.	2	74.6
3.	3	65.8
4.	4	58.7
5.	5	53.1
6.	6	48.2

Length of the given wire $=42.0 \mathrm{~cm}$

Diameter, measured using screw guage is $\mathrm{d}=0.22 \mathrm{~mm}$

Show Answer

Solution:

Since the balancing length $(\ell)$ is measured from the side of left gap, where $\mathrm{X}$ is connected, the formula used for determining $\mathrm{X}$ is:

$$ X=\left(\dfrac{\ell}{100-\ell}\right) \mathrm{R} $$

The above table can be completed, with the values of $X$ in each observation, as shown.

S1. No.	$\mathrm{R}$ $(\Omega)$	$\ell$ $(\mathrm{cm})$	$(100-\ell)$ $(\mathrm{cm})$	$\mathrm{X}=\frac{\mathrm{R} \ell}{100-\ell}$ $(\Omega)$
1.	1	85.7	14.3	5.99
2.	2	74.6	25.4	5.87
3.	3	65.8	34.2	5.77
4.	4	58.7	41.3	5.68
5.	5	53.1	46.9	5.66
6.	6	48.2	51.8	5.58

Mean $\mathrm{X}=5.76 \Omega$

Mean $\mathrm{X}=5.76 \Omega$

The formula for $\rho$ is

$\rho=\dfrac{\mathrm{X}}{\mathrm{L}}\left(\dfrac{\pi \mathrm{d}^{2}}{4}\right)$

$\mathrm{X}=5.76 \Omega$

$\mathrm{d}=0.22 \mathrm{~mm}=0.22 \times 10^{-3} \mathrm{~m}$

$\mathrm{L}=42.0 \mathrm{~cm}=0.42 \mathrm{~m}$

$\therefore \quad \rho=\dfrac{5.76 \Omega}{0.42 \mathrm{~m}} \times \dfrac{22}{7} \times \dfrac{\left(0.22 \times 10^{-3} \mathrm{~m}\right)^{2}}{4}$

$$ =5.22 \times 10^{-7} \Omega-\mathrm{m} $$

EXPERIMENT-12

Resistance of a given wire using Ohm’s law.

CONCEPT MAP

Ohm’s Law

Ohm’s law, for a conductor, states taht the strength of current (I), flowing through that conductor, varies directly as the potential difference (V) applied across its ends, provided all the physical conditions of that conductor, like temperature, length and cross-sectional area, remain unaltered.

That is, $I \propto V \quad$ (When the physical conditions of the conductor are kept constant).

or $\quad \mathrm{V} \propto \mathrm{I}$

$\therefore \mathrm{V}=\mathrm{RI}$

where $\mathrm{R}$ is the constant of proportionality; it is called the resistance of that conductor.

1. Resistance (R) of a conductor may be regarded as the obstruction offered by the conductor against the flow of charge carriers through it.

2. The cause of resistance is the collision between the moving charge carriers with the atoms / molecules of the conductor as well as with other moving charge carriers.

3. More is the chance of such collisions, more will be the resistance offered.

4. During collisions, energy of the moving charge carriers will be lost to atoms of the conductor. As these atoms / molecules get more energy the conductor, as a whole, will get heated up (or the temperature of the conductor will increase).

5. Thus flowing charges (or current) cause heating effect in the conductor.

6. Resistance (R) of a conductor depends on the following factors:

(i) nature of the conductor

(ii) length of the conductor $(\ell)$

and (iii) area of cross-section $\left(\mathrm{A}=\pi \mathrm{r}^{2}=\dfrac{\pi \mathrm{d}^{2}}{4}\right)$

7. It is found that:

(i) $\mathrm{R} \propto \ell \quad(\ell=$ length $)$

and (ii) $\mathrm{R} \propto \dfrac{1}{\mathrm{~A}} \quad(\mathrm{~A}=$ area of cross-section)

$\therefore \quad \mathrm{R} \propto \dfrac{\ell}{\mathrm{A}}$

or $\quad \mathrm{R}=\dfrac{\rho \ell}{\mathrm{A}}$

where $\rho$ is constant and it is called the resistivity or specific resistance of the material of the conductor. [ $\rho$ depends of the nature of the conductor; it also depends on the temperature of the conductor according to the formula $\rho _{(t)}=\rho _{(0)}(1+\alpha t)$, where $\alpha$ is called the temperature coefficient of resistivity.]

8. The unit of resistance (R) is volt-ampere ${ }^{-1}$ and is called ohm $(\Omega)$.

9. Since $\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}$ and $\rho$ depends on temperature, we can say that $\mathrm{R}$ also depends on temeprature. The resistance $(\mathrm{R})$ of a conductor varies according to the formula:

$R _{(t)}=R _{(0)}(1+\alpha t)$, where $\alpha$ is called temperature coefficient of resistnace, which is same as temperature coefficient of resistivity.

Ohmic and Non-ohmic Resistors

Materials, which offer resistance, are called resistors.

Ohm’s law is a conditional law, which is valid only when $R$ is kept constant by maintaining temperature, length and cross-sectional area as constant. Even if length and cross-section are kept un-changed, temperature can vary, due to heating cuased by the current flow. This variation in temperature can cuase variation in $\rho$, and thereby, variation in R. [For some alloys like mangamin and constantin, the value of $\alpha$ is almost zero and, therefore, the variation is $\rho$ and $\mathrm{R}$ is practically negligible for these materials. Hence resistance of wire made of such materials (for which $\alpha \simeq 0$ ) will remain practically as a constant even if their temperature changes. Such conductors, whose resistance (R) remains constant, are called ohmic conductors / resistors.

For them, $\mathrm{V}=\mathrm{RI}$ (where $\mathrm{R}=$ constant) is true for any temperature.

Therefore, $\mathrm{V} \propto \mathrm{I}$ or the $\mathrm{V}-\mathrm{I}$ graph for ohomic-conductor will be a straight line, as shown in the figure.

The slope of such a V-I graph will given the resistance ( $\mathrm{R}$ ) of the conductor, if $\mathrm{V}$ is plotted along $\mathrm{y}$-axis and $\mathrm{I}$ is taken along $\mathrm{X}$-axis.

The slope of I-V graph, therefore, will give the reciprocal of resistance, which is called the conductance $\left(\mathrm{C}=\dfrac{1}{\mathrm{R}}\right)$.

If, for a conductor / resistor, I does not always vary directly as V varies, then V-I graph cannot be a straight line; instead the graph will be a curve. Such conductors / resistors, for which V-I graph is not a straight line, are called non-ohmic resistors / conductors.

• It is to be noted that $\mathrm{V}=\mathrm{RI}$ formula is valid for non-ohmic resistors also, where the value of $\mathrm{R}$ is true only for that particular measured values of $\mathrm{V}$ and $\mathrm{I}$.
• For materials with a positive value of $\alpha$, (like tungston, nichrome etc.) $\mathrm{R}$ will increase with heating (or when I is passed for lnoger time). Their graphs will be like in figure (A).
• For materials with a negative value of $\alpha$, (like semiconductor and electrolytes), $\mathrm{R}$ will decrase with heating (or when I is passed for longer time). Their graphs will be similar to figure (B).

V-I Graph for Ohmic Conductors / Resistors

For ohmic resistors, the value of $\mathrm{R}$ is a constant for all values of $\mathrm{V}$ and $\mathrm{I}$. That is $\mathrm{I} \propto \mathrm{V}$. Hence, when V is doubled I will also be doubled and so on. Such a graph is shown here.

This is an ideal situation and in practical situations, it is difficult to obtain such a graph. This is because current I can cause heating of the resistor and thereby change its temperature; this, in turn, can change the value of R.

How to find Resistance from V-I Graph

Let the V-I graph for an ohmic resistor / conductor be as shown below.

The slope of this graph will be the ratio $\mathrm{V} / \mathrm{I}$ and, therefore, it will give the value of the resistance $\mathrm{R}$. To find the slope, we follow these steps:

(i) Two (quite far apart) points $\mathrm{A}$ and $\mathrm{B}$ are marked on the straight line graph.

(ii) Horizontal and verticals are drawn from both points, as shown.

(iii) The right angled triangle $\mathrm{ABC}$ is drawn.

(iv) Slope is calculated as:

$$ \begin{aligned} \text { Slope } & =\dfrac{\text { Perpendicular }}{\text { Base }} \\ & =\dfrac{B C}{A C} \\ & =\dfrac{V _{2}-V _{1}}{I _{2}-I _{1}} \end{aligned} $$

(v) The slope gives resistance $\mathrm{R}$.

$$ \therefore \quad \mathrm{R}=\dfrac{\mathrm{V} _{2}-\mathrm{V} _{1}}{\mathrm{I} _{2}-\mathrm{I} _{1}} $$

Note that, for a perfect ohmic resistor, $R=\frac{V _{1}}{I _{1}}$ or $R=\frac{V _{2}}{I _{2}}$ will also give the same value of $R$ as obtained from the slope.

In an experimental situation, the values of $\mathrm{V}$ and $\mathrm{I}$, measured by voltmeter and ammeter respectively, may not give the current values of $\mathrm{V}$ and $\mathrm{I}$ for the conductor / resistor, as there can be various errors creeping into the measurements. Here, in order to reduce the error in the calculation of R, we measure a numbers of values of $\mathrm{V}$ and their corresponding $\mathrm{I}$. We then plot the $\mathrm{V}-\mathrm{I}$ graph and obatin the most correct straight line (joining maximum points in the slot). The slope calculated will give the most correct value or $\mathrm{R}$.

Experimental Set-up and Procedure to Find R

The given resistance wire is connected is series with a current source, a key, a rheostat and an ammeter as shown here. A voltmeter is then connected across the resistance wire.

The rheostat can be adjusted to obtain different values of I. For every value of V (in volt) corresponding value of (in ampere) is measured and recorded. The V-I graph is plotted, with $\mathrm{V}$ along $\mathrm{y}$-axis and $\mathrm{I}$ along $\mathrm{x}$-axis. The slope of this V-I graph will give the resistance $\mathrm{R}$ of the given wire.

• Care should be takes to connect only the very ends of the resistor in the circuit.

• The least count, and range, of both the ammeter and the voltmeter, are to be recorded for reference.

• Parallax error must be avoided while taking the readings. The mirror strip, behind the scale of the meters, may be used for this.

• Heating effect should be reduced by switching off the circuit when not taking the readings.

• All terminals should be made tight.

EXPERIMENT-13

Potentiometer (i) Comparison of EMF off two primary cells. (ii) Determination of internal resistance of a cell.

CONCEPT MAP

(A)

(B)

Electro-chemical Cells

These are devices which generate electricity by converting chemical energy to electrical energy by means of chemical reactions.

An electro-chemical cell (simply called a cell) essentially consists of two electrodes, made of two different materials, dipped in a suitable electrolyte (taken in a container / vessel). For example, a simple voltaic cell consists of a copper rod (as positive electrode) and a zinc rod (as negative electrode) which are dipped in dilute sulphuric acid (taken as the electrolyte), taken in a glass beaker, as shown below.

$\mathrm{H} _{2} \mathrm{SO} _{4}$ moelcules dissociate in the solution to produce $\mathrm{H}^{+}$and $\mathrm{SO} _{4}^{–}$ions.

The higher electropositivity of copper will attract $\mathrm{H}^{+}$ions towards it, producing a positive potential for copper rod. Similarly the higher electronegativity of the zinc rod will attract $\mathrm{SO} _{4}^{–}$ions towards it, producing a negative potential for the zinc rod.

Thus a potential difference gets developed between the two electrodes. When the two electrodes are connected through an external circuit, electrons, deposited on the $\mathrm{Zn}-$ rod by $\mathrm{SO} _{4}^{–}$ions, wil flow through this circuit to the copper rod, producing a current in the circuit.

During this flow of electrons from $\mathrm{Zn}$-rod to the copper rod, through the external circuit, the following reactions take place:

(i) At the Zn-rod

$$ \mathrm{Zn}+\mathrm{SO} _{4}^{–} \rightarrow \mathrm{ZnSO} _{4}+2 \mathrm{e}^{-} $$

(These two electrons will flow to the copper rod through the external circuit).

(ii) At the $\mathrm{Cu}$-rod

The two $\mathrm{e}^{-}$, reaching the $\mathrm{Cu}-$ rod, will be absorbed by the $\mathrm{H}^{+}$ions.

$$ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H} _{2} $$

So, for every 2e- flowing through the external circuit one $\mathrm{H} _{2} \mathrm{SO} _{4}$ molcule will be consumed. Hence chemicals will get consumed when we get current flow in the external circuit.

• (Note that chemical reactions will begin only when the circuit of the cell is completed).

• (The $\mathrm{H} _{2}$ molecules, deposited on the copper-rod, can cause ‘polarisation’, becuase of which the flow of $\mathrm{e}^{-}$gets slowed down and eventually stops).

• The effect of polarisation can be reduced by oxidising hydrogen by using chemicals called ‘depolarisers’.

Daniel Cell and Leclanche Cell

These are electro-chemical cells in which depolarisers reduce the adverse effect of polarisation caused by deposition of hydrogen molecules at their positive electrode.

Daniel Cell

This is a modified simple-voltaic cell. $\mathrm{CuSO} _{4}$ is the chemical depolariser in Daniel cell. The construction of Daniel cell is shown in the diagram. The $\mathrm{Cu}$-vessel will itself act as the positive electrode. The negative electrode is an amalgamated $\mathrm{Zn}$-rod. $\mathrm{H} _{2} \mathrm{SO} _{4}$ is the electrolyte.

The chemical reactions, which take place at the $\mathrm{Zn}-$ rod (-ve electrode), are

$$ \mathrm{Zn}+\mathrm{SO} _{4}^{–} \rightarrow \mathrm{ZnSO} _{4}+2 \mathrm{e}^{-} $$

(These $2 \mathrm{e}^{-}$get deposited on the $\mathrm{Zn}$-rod. These reactions will continue to occur when the terminals of the cells are connected to the external circuit).

The $\mathrm{H}^{+}$, which are protons only, can easily go through the pores of the porcelain cylinder; they react with the $\mathrm{CuSO} _{4}$ solution.

$$ 2 \mathrm{H}^{+}+\mathrm{CuSO} _{4} \rightarrow \mathrm{H} _{2} \mathrm{SO} _{4}+\mathrm{Cu}^{++} $$

As $\mathrm{H} _{2}$ molecule is oxidised to $\mathrm{H} _{2} \mathrm{SO} _{4}$, deposition of $\mathrm{H} _{2}$ does not take place on the positive Cu electroode. The $\mathrm{Cu}^{++}$will receive the two $\mathrm{e}^{-}$from the $\mathrm{Cu}^{-}$electrode (or $\mathrm{Cu}^{-}$vessel) and get deposited on it as $\mathrm{Cu}^{-}$ atoms. (The potential difference (called EMF) developed across the positive and negative electrodes of Daniel cell is (nearly) 1.08 volt).

Leclanche Cell

There are two types of Lechanche cell:

(i) Wet Leclanche cell

(ii) Dry Leclanche cell

The “Dry Leclanche cell” is the “Dry cell” which we use commonly in the torches and ‘remotes’. The construction of “Wet Leclanche cell” normally called “Leclanche cell” is shown in the diagram below. $\mathrm{MnO} _{2}$ is used as depolariser in this cell. $\mathrm{MnO} _{2}$ is mixed with powdered charcoal to increase the conductivity. The EMF of this cell is 1.5 volt.

Primary and Secondary Cell

The electro-chemical cells are of two types:

(i) Primary cells

(ii) Secondary cells

Primary cells generate electricity by converting chemical energy into electrical energy. The chemical reactions taking place in primary cells, are irreversible. Hence they cannot be reacharged. Primary cells can not be re-used. Leclanche and Daniel cells are primary cells.

Secondary cells can store electrical energy as chemical energy and this energy gets converted back into electrical energy back when used. The chemical reactions are reversible. These cells can be recharged and therefore, re-used as well. They are also called accummulators as they accummulate energy during charging. Lead-acid accummulators (acid battery) used in automobile cars, invertors etc., are secondary cells.

EMF, Internal Resistance and Terminal Voltage of Cells

EMF refers to electro-motive-force, force which moves electrons. As we know, the force on the electrons cannot be measured directly; we measure the effect of this force in terms of ‘work done’ per unit charge. Every cell (source of electricity) has got an EMF, which depends on the nature of the electrodes used and the electrolyte of the cell.

Definitions of EMF

(i) EMF of a cell is defined as the work done, by the cell, in driving a unit charge through a circuit (closed) once, including the cell.

Hence, EMF is ‘work done’ by the cell per unit of charge.

Therefore, EMF, E $=\dfrac{\mathrm{W}}{\mathrm{q}}$

Hence the unit of EMF (E) is also $\mathrm{JC}^{-1}$ or volt. The above definition says that, if the EMF of cell is E ( $=2$ volt, say) it performs a work of 2 joules in driving one coulomb of charge through a closed circuit, including that cell.

(ii) The second definition of EMF is in terms of the potential difference. “EMF of a cell is defined as the potential difference between the terminals of the cell, when no current is drawn from it”.

[That is, EMF of a cell is its ’terminal voltage’ when current drawn from it is zero].

Definition of Internal Resistance (r)

It is the resistance offered, by the electrolyte of the cell, against flow of charges (i.e. ions) through it. Internal resistance depends on the following factors:

(i) the temperature of the electrolyte.

(ii) the nature of the electrolyte.

(iii) the concentration of the electrolyte.

(iv) the distance between the electrodes.

(v) the area of the electrodes dipped in the electrolyte.

Definition of Terminal Voltage

Terminal voltage $(\mathrm{V})$ is defined as the potential difference across the terminals of the cell, when some current is drawn from the cell. That is, $\mathrm{V}=$ potential difference when $\mathrm{I} \neq 0$.

Terminal voltage always appear across the external resistance, when $\mathrm{I} \neq 0$.

(Please note that $\mathrm{E}$ is also defined as potential difference across the terminals of the cell, but when $\mathrm{I}=0$ or $\mathrm{E}=\mathrm{V}$, when $\mathrm{I}=0$ ).

Relationship between $\mathrm{E}, \mathrm{r}$ and $\mathbf{V}$

Consider a simple circuit consisting of a cell of EMF E and internal resistance $r$, connected to an external resistance R. Let I be the current draw from the cell.

Here, the work done per unit charge, by the cell, is called the EMF.

i.e. $\mathrm{E}=$ work done per unit charge in driving one coulomb round the (closed) circuit.

or $\mathrm{E}=$ work done per unit change across $\mathrm{R}+$ work done per unit charge across $\mathrm{r}$

$\therefore \mathrm{E}=\mathrm{V}+v$ $\hspace {4 cm}$……..(1)

Here,

$\mathrm{V}=$ potential difference across $\mathrm{R}$

$v=$ potential difference across $\mathrm{r}$

(Remember that: potential difference is equal to work done per unit charge)

But Ohm’s law says that:

$\mathrm{V}=\mathrm{IR} \text { and } v=\mathrm{IR}$

$\therefore \quad \mathrm{E}=\mathrm{IR}+\mathrm{Ir}$

$\text { or } \quad E=I(R+r)$

$\text { or } \quad I=\dfrac{E}{R+r}$ $\hspace {3 cm}$……..(2)

Now, the terminal voltage of the cell is same as the voltage across the external resistance, that is equal to $\mathrm{V}=\mathrm{IR}$.

Therefore, from equation (1), we have

$\quad \quad \mathrm{V}=\mathrm{E}-\mathrm{v}$

$\text { or } \quad \mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\hspace {3 cm}$……..(3)

$\text { Now, } \mathrm{v}=\mathrm{E}-\mathrm{V}$

$\text { or } \quad \mathrm{Ir}=\mathrm{E}-\mathrm{V}$

$\text { or } \quad \dfrac{\mathrm{V}}{\mathrm{R}} \cdot \mathrm{r}=\mathrm{E}-\mathrm{V}$

$\therefore \mathrm{r}=\left(\dfrac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\hspace {3 cm}$……..(4)

Also, the terminal voltage, $V=I R$

$\therefore \quad V=\dfrac{E R}{R+r}$ $\hspace {3 cm}$……..(5)

Potentiometer

As the name suggests, it is a device which is used to measure potential difference. Since EMF is also measured as potential difference, a potentiometer can be use to measure EMF also.

A potentiometer essentially consists of a uniform resistance wire, of length from $4 \mathrm{~m}$ to $10 \mathrm{~m}$. This wire is fixed on a wooden board, in four to ten strands, each of length one metre, as shown in the figure. A metre scale is also fixed on this wooden board. The ends of this wire are provided with terminals A and B. (Usually we draw only four strands, for convenience).

Principle of Potentiometer

Let the resistance per unit length of the potentiometer wire $\mathrm{AB}$ be $\sigma \Omega \mathrm{cm}^{-1}$.

Suppose the wire AB of the wire is connected to a ‘primary’ circuit, consisting of a driver d.c source (a battery) of voltage $V _{D}$ in series with a rheostat $(\mathrm{Rh})$ and a key $(\mathrm{K})$. When the key is closed, the battery will drive a current, say I, through the wire AB. This current I, driven through the wire, will produce a uniform

‘Potential drop’ of I $\sigma$ volt per centimetre of the wire, all along the wire $\mathrm{AB}$. If the length of wire $\mathrm{AB}$ is $\mathrm{L}$ (which can be from $4 \mathrm{~m}$ to $10 \mathrm{~m}$ ) the potential difference across $\mathrm{AB}$ will be:

$\mathrm{V} _{\mathrm{AB}}=\mathrm{I} \sigma \mathrm{L}$ $\hspace {3 cm}$……..(1)

If A is at positive potential, the steady current, I, will produce a potential drop of I $\sigma$ for every one centimetre length of the wire.

Now if $\mathrm{J}$ is any point on the wire $\mathrm{AB}$, (which carries a steady current $\mathrm{I}$, driven by $\mathrm{V} _{\mathrm{D}}$ ), such that $\mathrm{AJ}=\ell$, then the potential drop across $\mathrm{AJ}(=\ell)$ length of this wire, is

$\mathrm{V} _{\mathrm{AJ}}=(\mathrm{I} \sigma) \ell$

or $\quad \mathrm{V}=\mathrm{I} \sigma \ell$ $\hspace {3 cm}$……..(2)

Here, $\mathrm{V} _{\mathrm{AJ}}=\mathrm{V}$, with A at higher $(+\mathrm{ve})$ potential and $\mathrm{J}$ at lower $(-\mathrm{ve})$ potential.

Thus, we see that $\mathrm{V} \propto \ell$, (as I $\sigma$ is a constant). This is the principle of the potentiometer.

Statement of the Principle of Potentiometer

If states that if a steady current (I) is passed through the uniform wire of the potentiometer (of resistance $\sigma \Omega \mathrm{cm}^{-1}$ ), then the potential drop (V) across any length ( $\ell$ ) of the wire is directly proportional to that length itselfi.e. $\mathrm{V} \propto \ell$.

Measurement of EMF of a Cell

Consider the potentiometer wire (AB) carrying a steady current (I), driven by the driver cell. Suppose a cell, of EMF (E), be connected to this potentiometer wire through a galvanometer and a jockey, as shown in the figure. (Care should be taken to ensure that positive terminals of both the driver battery and the cell are connected to the terminal A of the potentiometer).

The jockey can be moved the along the wire AB, till the galvanometer shows zero deflection. This is called the balancing of the cell across length $\mathrm{AJ}$ of the potentiometer wire.

Let $\mathrm{AJ}=\ell$ be called the balancing length, here.

Now, the two potential differences are balanced, in this case. These are:

(i) The potential difference across $\mathrm{AJ}(=\ell)$.

length of the potentiometere wire. (i.e. $\mathrm{V}=\mathrm{I} \sigma \ell$ )

(ii) The potential difference across the terminals of the cell.

Note that the cell is in a closed circuit here, but the net current drawn from this cell (ofEMF E) is zero (as the galvanometer shows zero deflection) at the balancing position $\mathrm{J}$ of the jockey. Thus, by definition of EMF, the potential difference available across the terminals of the cell is its EMF (E), (as no current is drawn from it).

Therefore, the $\operatorname{EMF}(\mathrm{F})$ of the cell is equal to the potential difference across AJ

or $\quad \mathrm{E}=\mathrm{V} _{\mathrm{AJ}}$

or $\quad \mathrm{E}=\mathrm{I} \sigma \ell$ $\hspace {3 cm}$……..(4)

By measuring the value of $I \sigma$ and $\ell$, we can find the value of $E$.

Potential Gradient along the Potentiometer Wire

From the principle of potentiometer, we have,

$\mathrm{V} \propto \ell$

$\therefore \quad \mathrm{V}=\mathrm{k} \ell$

or $\mathrm{k}=\frac{\mathrm{V}}{\ell}$

Here $\mathrm{k}$ is called the potential gradient. We have (since $\mathrm{V}=\mathrm{I} \sigma \ell$ )

$\mathrm{k}=\mathrm{I} \sigma$ $\hspace {4 cm}$……..(5)

or $\quad \mathrm{k} \propto \mathrm{I}$

It is clear, from equation (4) and (5), that if the value of I is less then the potential gradient will be less. Hence, to obtain the same potential difference $(\mathrm{V})$, we need a greater value of $\ell$ (when the potential gradient is lower. Thus a potentiometer, of lower potential gradient, is more accurate as the balancing length will be greater.

In the adjacent graph, $\mathrm{P}$ and $\mathrm{Q}$ represent plots for $\mathrm{V}$ against length, $\ell$, for two potentiometers. Clearly, A has lesser potential gradient and hence it would be preferred for the measurement of EMF as its accuracy is better.

Comparison of EMFs of two primary cells

Here, the aim is to compare the emfs of a Laclanche cell $\left(\mathrm{E} _{1}\right)$ and a Daniel cell $\left(\mathrm{E} _{2}\right)$.

Here, we use:

(a) a potentiometer wire $\mathrm{AB}$ of length (preferably) $10 \mathrm{~m}$; with $\sigma \Omega \mathrm{cm}^{-1}$ resistance.

(b) a primary circuit with a driver cell of voltage $\left(V _{D}\right)$ greater that the emf(E) of the two cells.

(c) The two cell whose EMFs are to be compared, along with a two-way key, galvanometer, a high resistance (H.R.) and a jockey.

(Note that the H.R. is used to protect the galvanometer from excess current damaging it. The H.R. may be short-circuited to get accurate balancing length).

All the above components are connected as shown in the circuit below.

Formula: It $\ell _{1}$ is the balancing length for Leclanche cell and $\ell _{2}$ is the balancing length for Daniel cell, then $\frac{\mathrm{E} _{1}}{\mathrm{E} _{2}}=\dfrac{\ell _{1}}{\ell _{2}}$

Experimental Set-up and Procedure

One the circuit is completed as shown in the figure above, the following steps are taken to check the correctness of the circuit.

• Ensure that all positive terminals ( $o f \mathrm{~V} _{\mathrm{D}}, \mathrm{E} _{1}$ and $\mathrm{E} _{2}$ ) are connected to terminal A of the potentiometer wire.

• The value of $\mathrm{V} _{\mathrm{D}}$ must be greater than the $\mathrm{EMF}\left(\mathrm{E} _{1}\right)$ of Leclanche cell $\left(\mathrm{E} _{1}=1.5 \mathrm{~V}\right)\left(\right.$ i.e. $\mathrm{V} _{\mathrm{D}}$ must be around 4V).

• When the Leclanche cell is included in the circuit (as is the case shown in the figure (above), we must obtain opposite deflections, in the galvanometer, when we press the jockey near A, on the wire, and then near B, on the wire.

[If opposite deflections are not obtained the rheostat can be adjusted in order to obtain opposite deflections]

• Once such opposite deflections are obtained, the circuit set-up is ready for use.

• Once the circuit is made to give opposite sides deflection, the balancing lengths for lechanche cell (of EMF $E _{1}$ ) and daniel cell (of EMF E ${ } _{2}$ ) are obtained, for the same steady current I (i.e. for the same rheostat position).

• Let $\ell _{1}$ be the balacing length measured for $\mathrm{E} _{1}$ and $\ell _{2}$ be the balancing length for $\mathrm{E} _{2}$. As $\mathrm{I} \sigma$ is the same for both $\ell _{1}$ and $\ell _{2}$ measurements, we can write:

$\mathrm{E} _{1}=\mathrm{I} \sigma \ell _{1} \text { amd } \mathrm{E} _{2}=\mathrm{I} \sigma \ell _{2}$

or $\dfrac{\mathrm{E} _{1}}{\mathrm{E} _{2}}=\frac{\ell _{1}}{\ell _{2}}$

• In order to take more observations, we can change the value of I, by adjusting the rheostat to a slightly different resistance value, keeping two important points in mind:

(a) $\ell _{1}$ and $\ell _{2}$ are obtained for same position of rheostat (or same current I) for every set of observations.

(b) opposite deflections are obtained in the case of Lechanche cell in each set of observations.

• The value of high resistance does not affect the balance point position as the observations are taken when $\mathrm{i} _{\mathrm{g}}=0$.

Determination of Internal Resistance ( $r$ ) of a Primary Cell

The objective of the experiment is to determine the resistance offered by the electrolyte of the cell; called the internal resistance ( $\mathrm{r}$ ) of the cell.

Since the internal resistance depends on many factors (like the temperature and concentration of the electrolyte, the distance between the electrodes and the surface area of the electrodes dipped in electrolyte), the value of $r$ may not give consistent results.

It is to be noted that, to find $r$ of the primary cell, we use the formula $r=\left(\dfrac{E}{V}-1\right) R$, with respect to the circuit diagram given aside. (Here $\mathrm{E}$ is the emf of the primary cell, $\mathrm{V}$ is the terminal voltage across the cell when the cell drives a current ithrough the external resistance $\mathrm{R}$ ).

Remember that, in the formula $r=\left(\dfrac{E}{V}-1\right) R, V$ is the terminal voltage when $R$ is the external resistance and $V$ will vary with $R$. Accurate determination of $r$ is possible only if we obtain the values of $\mathrm{E}$ and $\mathrm{V}$ correctly. This is done with the help of potentiometer, by obtaining the balancing lengths corresponding to $\mathrm{E}$ and $\mathrm{V}$.

Circuit Diagram and Formula

Here we use:

(a) A potentiometer wire $\mathrm{AB}$ of length (preferably) 10 metre and resistance $\sigma \Omega \mathrm{cm}^{-1}$.

(b) A driver circuit consisting of a battery of voltage $V _{D}$ a rheostat and a key connected in series with the wire $A B$. (The + terminal of $V _{D}$ unit be connected to $A$ ).

(c) A primary cell (say, a Leclanche cell), of EMF ’ $E$ ’ and internal resistance ’ $r$ ’ (for which $r$ is to be found out) connected with a galvanometer and high resistance along with a jockey, as shown (+ve terminal of E should be connected to the terminalA).

(d) A resistance box, in which a known and variable resistance $\mathrm{R}$ can be put, is connected across the Leclanche cell, through a one-way key. This tertiary circuit, will enable us to obtain E and V from the Leclanche cell.

The complete circuit is shown below.

$\mathrm{K} _{1}$ is key in the primary (driver) circuit and $\mathrm{K} _{2}$ is the key in the tertiary circuit.

Theory and Procedure

Suppose key $\mathrm{K} _{1}$ only ( $\mathrm{K} _{2}$ open) is closed and the rheostat is properly adjusted so that the galvanomeer gives opposite deflections when the jockey $\mathrm{J}$ is touched at $\mathrm{A}$ and then at $\mathrm{B}$.

Here, let I be the steady current flowing through the potentiometer wire $\mathrm{AB}$. Now, if $\mathrm{AJ}=\ell _{1}$ is the balancing length obatined, then, $\mathrm{E}=\mathrm{I} \sigma \ell _{1}$ gives the $\mathrm{EMF}$ of the leclanche cell.

Suppose next, a resistance $\mathrm{R}$ is included in the resistance box, in the tertiary circuit, and the key $\mathrm{K} _{2}$ is also closed.

Now, if $A J^{\prime}=\ell _{2}$ is the balancing length obtained, then, $\mathrm{V}\left(=\mathrm{I} \sigma \ell _{2}\right)$ gives the terminal voltage $\mathrm{V}$ of the Leclanche cell corresponding to a resistance ’ $\mathrm{R}$ ’ across the cell.

Therefore $\dfrac{E}{V}=\frac{\ell _{1}}{\ell _{2}}$

We have the formula for $\mathrm{r}$

$r=\left(\dfrac{E}{V}-1\right) R$, which now becomes $r=\left(\dfrac{\ell _{1}}{\ell _{2}}-1\right) R$

Knowing $\ell _{1}, \ell _{2}$ and $\mathrm{R}$, the value of $\mathrm{r}$ can be calculated.

The observations can be repeated for various values of ’ $R$ ’ as well as for various positions of the rheostat.

[Note that for a set of observations of $\ell _{1}$ and $\ell _{2}$ the positions of the rheostat should remain undisturbed].

The following precautions may be taken to reduce the error in the observations while using a potentiometer.

• Balancing length $(\ell)$ is measured, properly, from the end $\mathrm{A}$, along the wire.

• Overheating should be avoided by switching off the circuit immediately after every observation.

• All terminals should be made tight.

• The jockey should not be dragged hard along the potentiometer wire to avoid any change in the cross-sectional area of the wire.

Example:

The following circuit was used by a student to determine the internal resistance of a Leclanch cell.

The rheostat was kept at the same position throughout and $\mathrm{R}$ was varied. The observations recorded by the student are given in the table below. What are the maximum and minimum values of $\mathrm{r}$ $?$

Sr. No.	$\mathrm{R}$ $(\pi)$	Balancing when $\mathrm{K} _{1}$ alone is closed $\mathrm{I} _{1}(\mathrm{~cm})$	Balancing when $\mathrm{K} _{1}$ and $\mathrm{K} _{2}$ are both closed $1 _{2}(\mathrm{~cm})$
1.	10	824.5	446.0
2.	15	824.5	456.0
3.	20	824.5	496.0
4.	25	824.5	557.0
5.	30	824.5	585.0
6.	35	824.5	591.0

Show Answer

Solution:

The formula for determination of internal resistance, using potentiometer, is:

$$ \begin{aligned} r & =\left(\dfrac{E}{V}-1\right) R \\ \text { or } \quad r & =\left(\dfrac{\ell _{1}}{\ell _{2}}-1\right) R \end{aligned} $$

Using this formula the calculated value of $\mathrm{r}$ in each of the above observation is shown in the table below.

Sr. No.	$\mathrm{R}$ $(\Omega)$	$l _{1}$ $(\mathrm{~cm})$	$l _{2}$ $(\mathrm{~cm})$	$\mathrm{r}=\left(\dfrac{l _{1}}{l _{2}}-1\right) \mathrm{R}$ $(\Omega)$
1.	10	824.5	446.0	8.5
2.	15	824.5	456.0	12.1
3.	20	824.5	496.0	13.2
4.	25	824.5	557.0	12.0
5.	30	824.5	585.0	12.3
6.	35	824.5	591.0	13.8

Here, the minimum value of $r=8.5 \Omega$ and the maximum value of $r=13.8 \Omega$

EXPERIMENT–14

Aim: To determine the resistance $(G)$ and figure of merit $(k)$ of a galvanometer by half-deflection method.

Galvanometer

It is device used to detect very small currents flowing in a circuit.A galvanometer is a very sensitive device.

Principle of Working of Galvanometer

It is based on the principle that a coil, suspended or pivoted in a uniform magnetic field, will experience a torque whe a current (I) is passed through the coil.The extent of its rotation, against an (internal) restoring torque, will be proportional to the current strength. The extent of rotation can be measured with the help of a pointer, attached to the coil, when the pointer deflects and moves over a scale, graduated in divisions. The scale is a linear scale and the pointer is made to coincide with the zero of the scale (when no current is flowing) which is marked at the centre of the scale. This centre-zero scale enable us to observe the direction of current also as the pointer can deflect in both the directions from its zero position (which is at the middle of the scale).

Resistance (G) of the Galvanometer

The coil of the galvanometer is made of a long thin insulated copper wire wound on a rectangular metallic non-magnetic light frame. Since the wire is very thin and long, the wire provides resistance, even though the wire is made of copper. This resistance, provided by the coil, is called the resistance (G) of the galvanometer. The value of G ranges from around $50 \Omega$ to $250 \Omega$, depending on the sensitity of the galvanometer.

Figure of Marit (k)

It gives an idea of how sensitive the galvanometer is detecting current. Figure of merit ( $\mathrm{k})$ is defined as the strength of current which is required to deflect the pointer by one division on the scale.

That is $\mathrm{k}=$ current required for unit division deflection in the galvanometer.

If $\theta$ is the number of division deflections of the pointer, when a current I is passed through the galvanometer, $\mathrm{k}$ can be written as

$$ \mathrm{K}=\dfrac{\mathrm{I}}{\theta} $$

Since the galvanometer is very sensitive, $\mathrm{I}$ is normally in micro ampere only. Hence the unit of $\mathrm{k}$ is $\mu \mathrm{A} /$ division. The formula $\mathrm{K}=\dfrac{\mathrm{I}}{\theta}$ is used only to define the quantity called figure of merit of the galvanometer. Once the figure of merit is experimentally found out the current (I) can be obtained by observing the deflection produced by that current ( $\mathrm{as} \mathrm{I}=\mathrm{k} \theta$ ).

(Note that this current is very small and only in $\mu \mathrm{A}$ ).

Circuit Diagram

The circuit diagram, used for the determination of G, consists of a d.c source of emf E, a series variable resistance $\mathrm{R}$ (or a resisance box), the given galvanometer of (unknown) resistance G, a shunt resistance $\mathrm{S}$ (another resistance box) and two keys, which are connected in the manner as shown in the figure.

Derivation of the Formula for $G$ in the terms of $R$ and $S$

Let $\mathrm{R}$ be the resistance included in the resistance box 1 . Suppose $\mathrm{K} _{1}$ is closed and $\mathrm{K} _{2}$ is kept open. A current, say $\mathrm{I} _{\mathrm{g}}$, will flow through the galvanometer (Ensure that a sufficiently high value of $\mathrm{R}$ is put by the box so that the deflection of pointer is within the scale of the galvanometer). If ’ $\theta$ ’ is the number of divisions of deflection obtained on the scale and ’ $\mathrm{k}$ ’ is the figure of merit of the galvanometer, then

$$ \mathrm{I} _{\mathrm{g}}=\mathrm{k} \theta \quad \text { and } \quad \mathrm{I} _{\mathrm{g}}=\dfrac{\mathrm{E}}{\mathrm{R}+\mathrm{G}} $$

or $\quad \dfrac{E}{R+G}=k \theta$ $\hspace {4 cm}$ ……(1)

Now, suppose $\mathrm{K} _{2}$ is also closed (without changing the value of R). Here $\mathrm{S}$ gets connected across G. SInce a part of the current passes now through $\mathrm{S}$, connected in parallel to $\mathrm{G}$, the deflection in the galvanometer will decrease. Suppose the value of $\mathrm{S}$ is so adjusted that the deflection in the galvanometer is reduced exactly to $\dfrac{\theta}{2}$.

Let the current through the galvanometer, now, be $\mathrm{I} _{\mathrm{g}}^{\prime}$ and let the main current from the source be I. (Note that I will be more than $\mathrm{I} _{\mathrm{g}}$, the main current when $\mathrm{K} _{2}$ was open).

From the parallel connection of G and S,

We have

$$ I _{g}^{\prime}=\left(\dfrac{S}{G+S}\right) I $$

Also, $\mathrm{I} _{\mathrm{g}}^{\prime}=\mathrm{k}\left(\dfrac{\theta}{2}\right)$

$\therefore \quad \mathrm{I}\left(\dfrac{\mathrm{S}}{\mathrm{G}+\mathrm{S}}\right)=\mathrm{k} \dfrac{\theta}{2}$ $\hspace {4 cm}$ ……(2)

The combined resistance with circuit, now is

$$ \mathrm{R} _{\mathrm{eff}}=\mathrm{R}+\dfrac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}} $$

$\therefore \quad \mathrm{I}=\dfrac{\mathrm{E}}{\mathrm{R} _{\text {eff }}}$

or $\quad I=\dfrac{E}{R+\dfrac{G S}{G+S}}$

Substituting this value of I in equation(2), we get

$$ \left(\dfrac{E}{R+\dfrac{G S}{G+S}}\right)\left(\dfrac{S}{G+S}\right)=k \dfrac{\theta}{2} $$

or $\quad\left(\dfrac{E(G+S)}{R(G+S)+G S}\right) \dfrac{S}{(G+S)}=k \dfrac{\theta}{2}$

or $\dfrac{\mathrm{ES}}{\mathrm{R}(\mathrm{G}+\mathrm{S})+\mathrm{GS}}=\mathrm{k} \dfrac{\theta}{2}$ $\hspace {4 cm}$ ……(3)

Dividing equation (1) by equation (3), we have

$$ \dfrac{\mathrm{R}(\mathrm{G}+\mathrm{S})+\mathrm{GS}}{(\mathrm{R}+\mathrm{G}) \mathrm{S}}=2 $$

or

$$ \mathrm{RG}+\mathrm{RS}+\mathrm{GS}=2 \mathrm{RS}+2 \mathrm{GS} $$

or $\quad \mathrm{G}(\mathrm{R}-\mathrm{S})=\mathrm{RS}$

or $\quad \mathrm{G}=\dfrac{\mathrm{RS}}{\mathrm{R}-\mathrm{S}}$ $\hspace {4 cm}$ ……(4)

Since $R$ and $S$ are known resistances, the value of $G$ can be obatined using equation (4).

Formula for $\mathbf{k}$

From equation (1), we can find the value of $k$ as

$\mathrm{k}=\dfrac{\mathrm{E}}{(\mathrm{R}+\mathrm{G}) \theta}$ $\hspace {4 cm}$ ……(5)

Experimental Set-up and Procedure

A resistance box (1) of range ( $0-10,000 \Omega)$ is used to provide the known variable resistance R. Another resistance box (2), of range $(0-500 \Omega$ ), is use to give known variable resistance $S$. The d.c source is a battery of EMF ’ $E$ ‘. The given galvanometer, of unknown resistance ’ $G$ ‘, is connected as shown in the circuit.

Normally, a galvanometer with 30 divisions on its scale, on either side of the zero at the centre, is used in the experiment.

To determine ’ $G$ ‘, we follow the steps described below:

• Take out a high resistance of $\mathrm{R}(\simeq 10,000 \Omega)$ from box (1)

• Close $K _{1}$ (keeping $K _{2}$ open). Some deflection ( $\theta$ ) will be obtained in the galvanometer.

• By suitably adjusting the value of $R$, a deflection of $\theta=30$ divisions, exact, is obtained. The value of $\mathrm{R}$ is noted for $\theta=30$.

• Now $\mathrm{K} _{2}$ is also closed and the resistance $\mathrm{S}$ in the box (2) is so adjusted that the deflection reduces to $\frac{\theta}{2}=15$ exact. The value of $S$, for deflection of $\dfrac{\theta}{2}=15$ division, is recorded.

• Using the value of $\mathrm{R}$ and $\mathrm{S}$, the value of $\mathrm{G}$ is calculated from the formula $\mathrm{G}=\dfrac{\mathrm{RS}}{\mathrm{R}-\mathrm{S}}$

• Observations can be repeated for $\theta=28$ division, 26 division, 24 division etc.

• The mean value $\mathrm{G}$ will give the (most correct) value of resistance of the galvanometer.

• Once the value of $\mathrm{G}$ is obtained, the value of $\mathrm{k}$ can be calculated using the formula.

$$ \mathrm{k}=\dfrac{\mathrm{E}}{(\mathrm{R}+\mathrm{G}) \theta} $$

Here, $\mathrm{E}$ in the emf of the battery used, $\mathrm{G}$ is the mean value obtained from above, $\mathrm{R}$ is the series resistance, in box (1) for a given value of $\theta$.

The following precuations may be observed, in order to reduce the error in the observations.

• The plug keys of the resistance boxes should be properly cleaned using sand paper

• All plug keys and terminals must be tight.

• The readings in the galvanometer $(\theta)$ must be taken without any parallax error.

Example:

With the help of the experimental setup shown in the circuit diagram, using a cell of EMF $2 \mathrm{~V}$, a student obatined the following data for the determination of $\mathrm{G}$ and $\mathrm{k}$ of the galvanometer using half-deflection method.

Sr. No.	$\theta$ (Division)	$\mathrm{R}$ $(\Omega)$	$\theta / 2$	$\mathrm{S}$ $(\Omega)$
1.	30	3800	15	60
2.	28	4050	14	61
3.	26	4470	13	62
4.	24	4750	12	63
5.	22	5115	11	64
6.	20	5670	10	67

Calculte the values of $\mathrm{G}$ and $\mathrm{k}$.

Show Answer

Solution:

The formula for $\mathrm{G}$ and $\mathrm{k}$ are

$$ G=\dfrac{R S}{R-S} $$

and $k=\dfrac{E}{(R+G) \theta}$

To find $\mathrm{k}$, the mean value of $\mathrm{G}$, and $\mathrm{E}=2 \mathrm{~V}$, are to be used.

The values of $\mathrm{G}$ and $\mathrm{k}$, calculated for each set of observations, are shown in the table below.

$$ \mathrm{E}=2 \mathrm{~V} $$

Sr. No.	$\theta$ $($ Div. $)$	$\mathrm{R}$ $(\Omega)$	$\theta / 2$ $($ Div. $)$	$\mathrm{S}$ $(\Omega)$	$\mathrm{G}=\dfrac{\mathrm{RS}}{\mathrm{R}-\mathrm{S}}$ $(\Omega)$	Mean $\mathrm{G}$ $(\Omega)$	$\mathrm{K}=\dfrac{\mathrm{E}}{(\mathrm{R}+\mathrm{G}) \theta}$ $(\mathrm{A} /$ div. $)$
1.	30	3800	15	60	60.96		$1.72 \times 10^{-5}$
2.	28	4050	14	61	61.93		$1.73 \times 10^{-5}$
3.	26	4470	13	62	62.87	63.70	$1.69 \times 10^{5}$
4.	24	4750	12	63	63.84		$1.73 \times 10^{-5}$
5.	22	5115	11	64	64.81		$1.75 \times 10^{-5}$
6.	20	5670	10	67	67.80		$1.74 \times 10^{5}$

Mean $\mathrm{G}=63.70 \Omega$

Mean $\mathrm{k}=1.72 \times 10^{-5}$ ampere $/$ division.

EXPERIMENT-15

Aim: To find the focal length o

(i) convex mirror

(ii) concave mirror and

(iii) convex lens

Using parallax method.

Focal Length (f)

For spherical mirrors (convex and concave mirrors), the focal length ( $\mathrm{f}$ ) is defined as the distance of the principal focus $(\mathrm{F})$ from the pole $(\mathrm{P})$ of the mirror.

Now, principal focus, of a spherical mirror, is the point on the principal axis through which (or from which) rays, which are parallel to the principal axis, pass (or appear to diverge from), after reflection from the mirror.

(i) Convex Mirror

A spherical mirror with its reflecting surface convex in shape is alled a convex mirror. The centre of its aperture is named as the pole. The distance of the centre of curvature of the spherical mirror from the pole is called radius of curvature (R). According to the new Cartesian sign conventions, f and R are positive for convex mirror.

(ii) $f \simeq \dfrac{R}{2}$ for spherical Mirrors

For small aperture and comparitively longer radius of curvature, the focal length ( $f$ ) and the radius of curvature $(\mathrm{R})$, are related as $\mathrm{R} \simeq 2 \mathrm{f}$.

A Ray Incident Normally on a Mirror will Retrace its Path

As per the laws of reflection of light, a ray incident normally on any mirror / surface gets reflected back along the same path.

For a spherical mirror, any ray passing through the centre of curvature (C), or any ray directed towards the centre of curvature, will be incident normally on the mirror. There for they will trace the path back along the incident ray.

Ray-diagram for Detemining the Focal Length of a Convex Mirror, Using a Convex Lens

To determine the focal length (f) os a convex mirror, using the method of parallax, we require a real image to get formed, even while using a convex mirror. For this purpose we use a convex lens, along with the convex mirror. An optical pin is used as the real object here.

Any two rays originating from a point on the pin (for example the tip of the pin, assumed to be on the principle axis of the mirror), will get converged by the lens. If these converging rays are made to get incident normally on the reflecting surface of the convex mirror, they will retrace their path and converge back at the position of the object itself. Thus we will be able to get a real image of the optical pin at the same place where the object itself is.

Conversely, is we obtain a real image of the optical pin at the location of the pin itself, (by adjusting the position of the convex mirror, placed after the convex lens), then we can infer that the rays are incident normally on the surface of the convex mirror.

This scenario is shown in the diagram below.

When the mirror is removed, these rays with meet on the right side of the lens, and produce a real image.

Theory

From the two ray-diagrams drawn, it is clear that the position of the image (I), when the convex mirror is removed will give the centre of curvature of the mirror. (We locate the position of I in the second case by using an image pin and removing the parallax between the image and this image pin).

As image I is formed at ’ $\mathrm{C}$ ’ of the mirror, $\mathrm{MC}$ will be equal to $\mathrm{R}$. Therefore, $\mathrm{f}\left(=\dfrac{\mathrm{R}}{2}\right)$ can be calculated.

Optical Bench

An optical bengh consists of two nickel plated metallic rods of length about $1.5 \mathrm{~m}$, fixed parallel to each other, on two heavy metallic bases which are provided with levelling screws. The parallel rods have millimeter scales engraved on them. The bench has three or four metallic uprights which can slide freely along the length of the optical bench and can be fixed in any position. There is also an arrangment for lateral movement to bring the needles / pins, mirror or lens, fixed on the uprights, in one straight line. An index mark is made at the base of each upright. The position of the upright is read against these marks. The needles / pins serving as object, or image, and the mirror / lens, are mounted on these uprights. These can be lowered or raised, moved sidewise and can be fixed in the desired position such that tip of the needles / pin, the pole of the mirror (or the optical centre of the lens) are at the same height and are with least of ‘index error’.

Optical Bench

Parallax Method used to Locate the Position of a Real Image

A real image is formed when two or more rays, originating from a point on the object, are made to actually meet at a point elsewhere, by using optical devices. A real image can be obtained on a screen. In this experiment, the real image, of the optical pin, is located by using the parallax method.

Parallax refers to relative change in position of anobject, with respect to another object, when viewed from different positions. If the relative positions of the objects are different when viewed from different points, we say that there is parallex between such objects. When there is parallax, the objects are at different positions with respect to the viewer.

Now, we can adjust the positions of these objects, with respect to each other, such that the parallax gets removed, or the objects remain at same positions with respect to each other, when viewed from different positions. They may remain coinciding for the viewer, at all positions of the view. This is called removing the parallax. When we remove parallax, between the tips of an image pin and the real image of an object pin, we say that the image (of the object pin) is at the same position as that of the image pin.

To remove parallax in optical experiments we use a refernce optical pin, often called image pin, which is fixed on an upright. When parallax is removed between the real image, and this image pin, the position of the real image is same as that of the image pin and can be noted from the index line of its upright.

Parallax can be removed between the real image of an object pin, and the pin itself also, as is case in the first part of this experiment. While removing the parallex between a real image and the reference pin (image pin), it is advised to observe their relative position using only one eye (keeping the other eye closed).

Procedure for Determining the Focal Length of the Convex Mirror

• Mount the convex mirror on a holder and then fix the holder on the upright of the optical bench. Simiarly mount a convex lens and fix it on another upright.

• Place the upright, holding the convex lens, at around $75 \mathrm{~cm}$ marking on the bench.

• Place the convex mirror beyond this lens and at around $100 \mathrm{~cm}$ marking.

• Now place an object pin $(\mathrm{O})$ on an upright and then place this upright in front of the lens, so that rays from the object pin pass through the lens and fall on the convex mirror placed beyond it.

• Viewing from the same side of the object pin, in the direction of the lens, keeping a minimum distance of $25 \mathrm{~cm}$ between the eye and the object pin, we will be able to see an inverted real image of the object pin, formed by the convex lens and the convex mirror combination, on the same side of the lens as the object is.

• Now, remove parallex between this real image and the object pin itself by adjusting the position of the convex mirror. (Ensure that the tip of the image coincides with the tip of the object all the while, during the adjustments we do for the removal of parallax).

• Once the parallax is removed between the object pin and its image, (formed by the lens and mirror together), we can be sure that the rays, passing through the lens, are falling normally on the convex mirror and so they are getting retraced back, to produce the real image back at the object itself.

• Record the positions of the object pin, lens and the mirror as $\mathrm{O}, \mathrm{L}$ and $\mathrm{M}$.

• Now remove the mirror and place another optic pin, a reference pin (called image pin) on another upright.

• View the real image, of the object pin, formed by the lens, from the other side of the lens.

• Make the tips of this real image of the object pin, and the image pin coincide. Adjust the positions of the image pin to remove the parallax, so as to obtain the position of the real image (I) formed by the lens alone.

(Here we are getting the situation corresponding to the second part of the ray-diagram.)

• Record the position of I.

Calculation of $f$

Now, the distance MI will give the radius of curvature of the convex mirror.

Therefore, $\mathrm{MI}=\mathrm{R}$

But $\quad f=\dfrac{R}{2}$

$\therefore \mathrm{f}=\dfrac{\mathrm{MI}}{2}$

(To take more observations the position of the object pin can be varied, keeping the position of lens the same. Accordingly the positions of mirror (M) and image (I) will change. But the distance $\mathrm{MI}=\mathrm{R}$ will remain the same, within experimental error).

To find the Focal Length of a Concave Mirror

Concave Mirror

It is a spherical mirror whose reflecting surface is concave in shape. A concave mirror can form real images,on its own, if the object is kept beyond its principal focus (F).

Ray-diagram

Here, object pin is placed at $\mathrm{O}$, beyond $\mathrm{C}$. Therefore the image I will be formed between C and F. Also,

$$ \begin{aligned} & \mathrm{OP}=\mathrm{u}(\text { object distance }) \\ & \mathrm{IP}=\mathrm{v}(\text { image distance }) \end{aligned} $$

Theory

(a) Once we have obtained the values of $u$ and $v$, we can find the focal length f using the formula.

$$ \frac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} $$

Note that both $u$ and $v$ are negative according to the cartesian sign convention, as both are measured to the left of $\mathrm{P}$ (for real images). Thus $\mathrm{f}$ is also obtained as negative.

(b) After we have obtained many values of $v$ corresponding to different values of $u$, we can plot a graph of $u$ against $v(u-v$ graph). (Note that the scales selected for $u$ and $v$ axis should be the same and the origin should be taken at zero for both the axes). Since both $u$ and $v$ are negative, the $u-v$ graph will be drawn in the third quadrant.

After plotting the $\mathrm{u}-\mathrm{v}$ graph, $45^{\circ}$ angle bisector is drawn from the origin to interesect the graph at $\mathrm{O}^{\prime}$. The co-ordinates or the $u$ and $v$ values of $\mathrm{O}^{\prime}$, are $\mathrm{u}=\mathrm{OA}$ and $\mathrm{v}=\mathrm{OB}$.

But from the geometry, $\mathrm{OA}=\mathrm{OB}$

$\therefore \quad \mathrm{u}=\mathrm{v}$

This means that object and image are at same distance from the pole of the mirror. This happens when they are at $\mathrm{C}$ that is $\mathrm{OA}=\mathrm{OB}=\mathrm{R}=2 \mathrm{f}$.

Therefore, $\mathrm{f}=\dfrac{\mathrm{OA}}{2}=\dfrac{\mathrm{OB}}{2}$

The average of $\dfrac{\mathrm{OA}}{2}$ and $\dfrac{\mathrm{OB}}{2}$ will give the mean value of focal length.

$$ \because \quad \mathrm{f}=\dfrac{\mathrm{OA}+\mathrm{OB}}{4} $$

(c) Corresponding to the various values of $u$ and $v$, corresponding values of $\dfrac{1}{u}$ and $\dfrac{1}{v}$ are calculated and $\mathrm{a}\left(\dfrac{1}{\mathrm{u}}\right) \mathrm{vs}\left(\dfrac{1}{\mathrm{v}}\right)$ graph is drawn. (Here also same scales are selected for both the axes and origin is taken as zero). The graph is a straight line graph; it is extrapolated to obtain the $\mathrm{x}$ and $\mathrm{y}$-intercepts, as shown.

From the graph, it is clear that the co-ordinates of $\mathrm{C}$ are $\left(\dfrac{1}{\mathrm{u}}, 0\right)$, which means that $\dfrac{1}{\mathrm{v}}$ is 0 or $\mathrm{v}$ is tending to infinity for point $\mathrm{C}$. When $\mathrm{v} \Rightarrow \infty, \mathrm{u}=\mathrm{f}$.

Therefore $\dfrac{1}{\mathrm{u}}=\dfrac{1}{\mathrm{f}}$, for point $\mathrm{C}$.

or $\dfrac{1}{\mathrm{f}}=\mathrm{OC}$

$\therefore \quad \mathrm{f}=\dfrac{1}{\mathrm{OC}}$

Similarly, $\mathrm{f}=\dfrac{1}{\mathrm{OD}}$

The mean of $\dfrac{1}{\mathrm{OC}}$ and $\dfrac{1}{\mathrm{OD}}$ will give the correct value off. (Note that all quantities are-ve)

Experimental Set-up and Procedure

Here also we use the optical bench, with the uprights, as described in case of the convex mirror experiment. The concave mirror is mounted on the upright and this upright is kept at, say, $100 \mathrm{~cm}$ marking on the bench.

The object pin is placed at a point which is beyond $\mathrm{C}$ for the mirror, so that the real image formed is smaller in size.

Now, using an image pin as reference pin, the parallax, between the image of the object pin and the image pin, is removed to locate the position of the image, accurately. Once the image is located, using the imagepin, the positions of object pin ’ $\mathrm{O}$ ‘, mirror ’ $\mathrm{M}$ ’ and image ’ $\mathrm{I}$ ’ are recorded.

The whole procedure is repeated for different positions of object, each time recording the positions of $\mathrm{O}$, $\mathrm{M}$ and $\mathrm{I}$.

In each case:

$$ \begin{aligned} & \mathrm{u}=\mathrm{OM} \text { and } \\ & \mathrm{v}=\mathrm{IM} \end{aligned} $$

The focal length ( $\mathrm{f}$ ) is then calculated using:

(a) Formula $\left(\dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}+\dfrac{1}{\mathrm{u}}\right)$ (b) (u) vs (v) graph

(c) $\left(\dfrac{1}{\mathrm{u}}\right) \mathrm{vs}\left(\dfrac{1}{\mathrm{v}}\right)$ graph, as explained in the theory.

(Note that when $\mathrm{u}>2 \mathrm{f}$, the image is smaller and when $\mathrm{u}<2 \mathrm{f}$, the image is bigger than the object. Care should be taken to have values of $\mathrm{u}$ where $\mathrm{u}>2 \mathrm{f}$ and also where $\mathrm{u}<2 \mathrm{f}$, in order to obtain proper plots of the graphs).

To find the Focal Length of a Convex Lens

Convex Lens

It is a piece of transparent medium bound by two spherical convex surfaces (bi-convex lens).

The centre of the lens is known as optical centre and it is marked as $\mathrm{P}$.

Focal length ( $\mathrm{f}$ ) is the distance of the principal focus $(\mathrm{F})$ from the optical centre $(\mathrm{P})$.

i.e. $\mathrm{PF}=\mathrm{f}$

The principal focus $(\mathrm{F})$ is the point on the principal axis at which rays, parallel to the principal axis, pass after refraction through the lens.

Focal lengh of a convex lens is positive.

Ray-diagram

The convex lens can form real images when the object is kept beyond its $\mathrm{F}$ point. If the object is beyond $2 \mathrm{~F}$, the image is smaller in size and if it is between $\mathrm{F}$ and $2 \mathrm{~F}$, the image is bigger than the object.

The position of the image (I) is located by removing the parallax between the image pin (reference pin) and the image of the object.

Once the image (I) is located we have $\mathrm{OP}=\mathrm{u}$ and $\mathrm{PI}=\mathrm{v}$. Also we can have many positions of object $(\mathrm{O})$ and corresponding positions of image (I) to get various values of $u$ and $v$.

Theory

(a) $\operatorname{IF~} \mathrm{PO}(=\mathrm{u})$ and $\mathrm{PI}(=\mathrm{v})$ are obtained, we can find the focal length (f) using the formula:

$$ \dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}-\dfrac{1}{\mathrm{u}} $$

(Here $u$ is negative and $v$ is positive according to sign convention. Therefore $f$ of convex lens is positive).

(b) We can plot the u-v graph (in second quadrant) and obtain focal length (f). For plotting (u) vs (v) graph we should use same scale for both ’ $u$ ’ and ’ $v$ ’ axes and the origin should be at zero.

Once the graph is drawn, drop the perpendicular and draw the horizontal, as shown.

From the geometry if is clear that $|\mathrm{OA}|=\mathrm{OB}$ or $|\mathrm{u}|=\mathrm{v}$. Therefore $|\mathrm{OA}|=\mathrm{OB}=2 \mathrm{f}$.

Hence $\quad f=\dfrac{|\mathrm{OA}|}{2}=\dfrac{\mathrm{OB}}{2}$

or $\mathrm{f}=\dfrac{|\mathrm{OA}|+\mathrm{OB}}{4}$

(c) We can also find the values of $\dfrac{1}{u}$ and $\dfrac{1}{v}$ and plot a $\left(\dfrac{1}{u}\right) v s\left(\dfrac{1}{v}\right)$ graph. It will be a straight line.

The value of ’ $f$ ’ can be obtained from this graph as:

$$ \dfrac{1}{\mathrm{f}}=\dfrac{\left|\dfrac{1}{\mathrm{OC}}\right|+\dfrac{1}{\mathrm{OD}}}{2} $$

Experimental Set-up and Procedure

The optical bench is used here also. The convex lens is mounted on the upright and it is placed at (around) the $75 \mathrm{~cm}$ mark on the bench.

The object pin is placed at a point which is beyond $2 \mathrm{~F}$ for the lens, so that the real image formed, on the other side of the lens is smaller in size.

Now using an image pin, as reference pin, the location of this real image is obtained by removing the parallax between the two.

Once the image is located, the positions of the object pin ’ $\mathrm{O}$ ‘, the lens ’ $\mathrm{L}$ ’ and the image ’ $\mathrm{I}$ ’ are recorded. Here $\mathrm{LO}=|\mathrm{u}|$ and $\mathrm{LI}=\mathrm{v}$.

The procedure is repeated to obtain many values for $v$ corresponding to different values of $u$.

Calculations of $f$

(a) From the recorded values of $u$ and $v$ for each set, $f$ is calculated from the formula

$$ \dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}-\dfrac{1}{\mathrm{u}} $$

(b) From the (u) vs (v) graph.

$$ \mathrm{f}=\dfrac{|\mathrm{OA}|+\mathrm{OB}}{4} $$

(c) From the $\left(\dfrac{1}{u}\right) v\left(\dfrac{1}{v}\right)$ graph

$$ \mathrm{f}=\dfrac{\left|\dfrac{1}{\mathrm{OC}}\right|+\dfrac{1}{\mathrm{OD}}}{2} $$

Example:

Astudent, performing the experiment to find the focal length of convex mirror, took the following two sets of observations. Obtain the focal length of the convex mirror.

Object-pin position (cm)	Lens positions $(\mathrm{cm})$	Convex Mirror positions when the Image of the object-pin is formed at the object-pin it self $(\mathrm{cm})$	Image position when the convex Mirror is removed $(\mathrm{cm})$
70.5	50.0	42.0	31.4
64.0	50.0	35.5	15.0

Show Answer

Solution:

The ray diagram for the first set of observation is shown below.

From this, clearly, $\mathrm{R}=(42.0-31.4) \mathrm{cm}=10.6 \mathrm{~cm}$

$$ \therefore \quad \mathrm{f}=\dfrac{\mathrm{R}}{2}=5.3 \mathrm{~cm} $$

From the second set, $\mathrm{R}=(35.5-15.0) \mathrm{cm}=10.5 \mathrm{~cm}$

$$ \therefore \quad \mathrm{f}=\dfrac{\mathrm{R}}{2}=5.25 \mathrm{~cm} $$

Mean $\mathrm{f}=\dfrac{5.3+5.25}{2}=5.3 \mathrm{~cm}$

Example:

The following observations were recorded by a student performing concave mirror experiment on optical bench, using parallax method.

Sr. No.	Position of Mirror $(\mathrm{cm})$ $\mathrm{M}$	Position of Object $(\mathrm{cm})$ O	Position of Image $(\mathrm{cm})$ I
1.	0.0	20.0	34.0
2.	0.0	17.0	46.4
3.	0.0	15.0	56.8
4.	0.0	35.0	20.7
5.	0.0	38.0	19.2
6.	0.0	41.0	17.6
7.	0.0	44.0	16.1

Calculate the focal length (f) of the concave mirror from these observation by:

(a) calculation using the formula $\dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}+\dfrac{1}{\mathrm{u}}$

(b) by plotting (u) vs (v) graph.

(c) by plotting $\left(\dfrac{1}{\mathrm{u}}\right) \mathrm{vs}\left(\dfrac{1}{\mathrm{v}}\right)$ graph.

Show Answer

Solution:

In each observation we find the value of $u$ and $v$ as:

$$ \begin{aligned} & |\mathrm{u}|=\mathrm{MO} \\ & |\mathrm{v}|=\mathrm{MI} \end{aligned} $$

Also we calculate the values of $\dfrac{1}{\mathrm{u}}$ and $\dfrac{1}{\mathrm{v}}$ for each set.

Finally we calculate $f=\dfrac{u v}{u+v}$ for each set of observations.

The completed table is shows below:

Sr. No.	Position of Mirror $\mathrm{M}(\mathrm{cm})$	Position of Object $\mathrm{O}(\mathrm{cm})$	Position of Image $\mathrm{I}(\mathrm{cm})$	$-u$ $(\mathrm{~cm})$	$-v$ $(\mathrm{~cm})$	$-\dfrac{1}{u}$ $\left(\mathrm{~cm}^{-1}\right)$	$-\dfrac{1}{v}$ $\left(\mathrm{~cm}^{-1}\right)$	$f=\dfrac{u v}{u+v}$ $(\mathrm{~cm})$
1.	0.0	20.0	34.0	20.0	34.0	0.05	0.029	12.6
2.	0.0	17.0	46.4	17.0	46.4	0.059	0.022	12.4
3.	0.0	15.0	56.8	15.0	56.8	0.067	0.018	11.9
4.	0.0	35.0	20.7	35.0	20.7	0.029	0.048	14.3
5.	0.0	38.0	19.2	38.0	19.2	0.026	0.052	12.8
6.	0.0	41.0	17.6	41.0	17.6	0.024	0.057	12.3
7.	0.0	44.0	16.1	44.0	16.1	0.013	0.063	11.8

(i) Meanf(fromcalculation), using $f=\dfrac{-u v}{u+v}$

$$ \begin{aligned} \mathrm{f}= & -\left(\dfrac{12.6+12.4+11.9+12.8+12.3+11.8}{6}\right) \mathrm{cm} \\ & =-12.3 \mathrm{~cm} \end{aligned} $$

(ii) (u) vs (v) graph

$$ \mathrm{f}=-13.2 \mathrm{~cm} $$

(iii) $\left(\dfrac{1}{\mathrm{u}}\right) \mathrm{vs}\left(\dfrac{1}{\mathrm{v}}\right)$ graph $\mathrm{f}=-12.8 \mathrm{~cm}$

EXPERIMENT–16

Plot of angle of deviation (<D) versus angle of incidence (<i) for a triangular prism.

CONCEPT MAP

Measurement of the Refracting Angle of the Prism

The refracting angle, of a given prism, can be measured by finding the positions of the reflected rays, corresponding to the incident rays $\left(\mathrm{PP} _{1} \mathrm{P} _{2} \mathrm{QD}\right)$ and $\left(\mathrm{P}^{\prime} \mathrm{P} _{1}{ }^{\prime} \mathrm{P} _{2}{ }^{\prime} \mathrm{Q}\right.$ ‘E), two incident rays taken along the perpendicular to the base $\mathrm{BC}$ of the prism.

It can be shown that

$$ <\mathrm{RSR}^{\prime}=2(<\mathrm{A}) $$

Tracing the Path of Rays through a Prism

The set up used for tracing the path, of a given incident rays through a prism is as shown. The accompyning diagram shows how the different angles can be measured.

Refraction of Light through a Prism

The ray diagram, showing the refraction of (monochromatic) ray of light by a prism, is given below.

In this diagram

$<\mathrm{i}$ and $<\mathrm{r}=$ angles of incidence and refraction at the face $\mathrm{AB}$

$<\mathrm{r}$ ’ and $<\mathrm{e}=$ angles of incidence and refraction at the face $\mathrm{AC}$

$[<e$ is also known as the angle of emergence].

$<\mathrm{D}=$ angle of deviation

$<\mathrm{A}=$ angle of the prism.

These angles are related as follows:

$<\mathrm{r}+<\mathrm{r}^{\prime}=<\mathrm{A}$

and $<\mathrm{i}+<\mathrm{e}=<\mathrm{A}+<\mathrm{D}$

Graph between $<$ i and $<$ D

The dependence of the angle of deviation, on the angle of incidence, is shown in the graph given here.

There is one, and only one, angle of incidence $\left(<\mathrm{i}=<\mathrm{i} _{\mathrm{m}}\right)$, at which the angle of deviation $(<\mathrm{D})$ has its minimum value $\left(=<\mathrm{D} _{\mathrm{m}}\right)$.

However, there can be two values of the angle of incidence (one greater than $\mathrm{i} _{\mathrm{m}}$ and the other less than $\mathrm{i} _{\mathrm{m}}$ ) for which the angle of deviation has (some other) same value.

Prism

A prism may be regarded as a section of a transparent medium, bounded by two plane refracting surfaces, meeting at an angle.

Angle of the Prism

The angle between the two ‘faces’ (surface ABEF and ACGE) is known as the angle of the prism ( $<\mathrm{A})$. The ‘base’ of the prism is the retangular surface (BCGF), at its bottom.

Refracting Edge a Prism

The two plane refracting surfaces, of a prism, meet along the line $\mathrm{AE}$. We call this line $(\mathrm{AE})$ as the ‘refracting edge’ of a prism.

Principal Section of a Prism

Let a plane, perpendicular to its refracting edge, be used to obtain a ‘section’ of the prism. We refer to such a section as the principal section of the given prism.

Example:

The angle of deviation for a given prism, (made from a material of refractive index $\mu$ ), has the same value for two different values, say $i _{1}$, and $i _{2}$, for the angle of incidence. If the angles of emergence, for these the values of angle of incidence, are $e _{1}$ and $e _{2}$, respectively, we would have

(i) $\mathrm{i} _{1}-\mathrm{i} _{2}=\left(\mathrm{e} _{1}-\mathrm{e} _{2}\right)$

(ii) $\mathrm{i} _{1}-\mathrm{i} _{2}=\left(\mathrm{e} _{2}-\mathrm{e} _{1}\right)$

(iii) $\mathrm{i} _{1}+\mathrm{i} _{2}=\mu\left(\mathrm{e} _{2}+\mathrm{e} _{1}\right)$

(iv) $\left(\mathrm{i} _{1}+\mathrm{i} _{2}\right)=\mu\left(\mathrm{e} _{1}-\mathrm{e} _{2}\right)$

Solution:

We know that, for a prism,

$$ \mathrm{i}+\mathrm{e}=\mathrm{A}+\mathrm{D} $$

Here, as D has the same value for the two sets of angles, we have.

$$ \begin{aligned} & \mathrm{i} _{1}+\mathrm{e} _{1}=\mathrm{A}+\mathrm{D}=\mathrm{i} _{2}+\mathrm{e} _{2} \\ & \therefore\left(\mathrm{i} _{1}-\mathrm{i} _{2}\right)=\left(\mathrm{e} _{2}-\mathrm{e} _{1}\right) \end{aligned} $$

EXPERIMENT-17

Refractive index of a glass slab using a travelling microscope.

CONCEPT MAP

Refraction

There is a change in the speed of light as it goes from one medium to another. The phenomenon, associated with this change in speed of light, are referred to as ‘refraction of light’.

Refraction is usually observed through a change in the direction of propagation (or ‘bending’ of the path) of a ray of light as it goes (obliquely) from one medium to another.

Refractive Index

The refractive index $\left({ } _{1}^{2} \mu\right)$ of a medium (say medium 2), with respect to another medium (say medium 1), equals the ratio of the speed of light in medium $1\left(=\mathrm{v} _{1}\right)$ to its speed in medium $2\left(=\mathrm{v} _{2}\right)$

$$ { } _{1}^{2} \mu=\dfrac{\mathrm{v} _{1}}{\mathrm{v} _{2}} $$

We also define ${ } _{1}^{2} \mu$ through Snell’s law:

$$ { } _{1}^{2} \mu=\dfrac{\sin \mathrm{i}}{\sin \mathrm{r}} $$

Real Depth and Apparent Depth

‘Refraction’ causes a pond of water to appear shallower than it actually is. It can also cause a rectangular glass slab to appear less thick than it actually is.

We use the terms ‘apparent depth’ and ‘real depth’ in this context.

A relation between these two ‘depths’ can be obtained as follows. Consider a rectangular glass slab, $\mathrm{ABCD}$, of thickness, t, say.

Let there be a point $P$ at the bottom of the glass slab. A ray $\mathrm{PQ}$, incident at point $\mathrm{Q}$, bends away from the normal, and goes along $\mathrm{QR}$, during its propagation from glass to air. When produced backwards, it intersects a ’normally incident ray’, PNP’, at the point $\mathrm{P}^{\prime}$. $\mathrm{P}^{\prime}$ is, therefore, the virtual image of point $\mathrm{P}$ when this point is viewed from above the glass slab. The thickness of the glass slab, therefore, appears to be P’N ( $\therefore$ P’N = apparent thickness) when its actual thickness is PN ( $\therefore \mathrm{PN}=$ real thickness).

Now, from Snell’s law,

$$ \begin{aligned} & { } _{\text {glass }}^{\text {air }} \mu=\dfrac{\sin \mathrm{i}}{\sin \mathrm{r}} \\ & \text { Again } \sin \mathrm{i}=\dfrac{\mathrm{PQ}^{\prime}}{\mathrm{PQ}} ; \sin \mathrm{r}=\cos \left(<\mathrm{P}^{\prime} \mathrm{QN}\right)=\dfrac{\mathrm{NQ}}{\mathrm{P}^{\prime} \mathrm{Q}} \\ & \therefore \dfrac{\sin \mathrm{i}}{\sin \mathrm{r}}=\dfrac{\mathrm{PQ}^{\prime}}{\mathrm{PQ}} \times \dfrac{\mathrm{P}^{\prime} \mathrm{Q}}{\mathrm{NQ}}=\dfrac{\mathrm{P}^{\prime} \mathrm{Q}}{\mathrm{PQ}} \quad\left(\because \mathrm{PQ}^{\prime}=\mathrm{NQ}\right) \\ & \approx \dfrac{\mathrm{P}^{\prime} \mathrm{N}}{\mathrm{PN}} \text { (Assuming } \mathrm{Q} \text { to be very near to } \mathrm{A} \text { ) } \\ & \therefore \quad \underset{\text { glass }}{\text { air }} \mu \simeq \dfrac{\mathrm{PN}}{\mathrm{P}^{\prime} \mathrm{N}}=\dfrac{\text { Realdepth }}{\text { Apparent depth }} \end{aligned} $$

Travelling Microscope

We can use the special microscope, called a travelling microscope, for finding the real and apparent thicknesses of a given glass slab. The ‘stands’, attached to the tube of the ‘compound microscope’ used here, are such that the microscope tube can be moved both ‘up and down’ (i.e. vertically) as well as ‘right and left’ (i.e. horizontally). There are ‘scales’ (‘main scale’, with its attached ‘vernier scale’) that enable one to measure the (vertical or horizontal) movement of the ‘cross-wires’ in the ’eye-piece’ of this microscope. The ’least count’, of the vernier used here is generally $0.0005 \mathrm{~cm}$ or $0.001 \mathrm{~cm}$.

(Travelling Microscope)

To find the apparent depth, of the given glass slab, we put a sharp dark ink mark on a white sheet of paper and focus the microscope on it. Its ‘scale reading’ is taken. The glass slab is now put on top of the mark and the microscope tube is ‘raised up’ so that the image (of the mark) again comes in ‘sharp focus’ of the microscope. The ‘scale reading’ of the microscope is again taken. The difference, between this reading, and the first reading, gives the ‘apparent depth’ of the given glass slab.

To find the ‘real depth’, some lycopodium power, or fine chalk dust, is sprinkled on top of the glass slab. The microscope tube is then raised further till this powder (or dust) particles come in ‘sharp focus’ of the microscope. The scale reading of the microscope is again taken. The difference between this (third) reading, and the first reading, tives the ‘real depth’ of the glass slab.

We then have

$$ \underset{\text { glass }}{\operatorname{air}} \mu={ } _{\mathrm{a}}^{\mathrm{g}} \mu=\left(\dfrac{\text { Real depth }}{\text { Apparent depth }}\right) $$

Example:

Three different liquids are filled in three beakers to heights $h _{1}, h _{2}, h _{3}$, (respectively) that are in the inverse ratio of their respective refractive indices, $\mu _{1}, \mu _{2}, \mu _{3}$.

The ratio of their apparent heights, when viewed from above, would be

(i) $1: 1: 1$

(ii) $\left(\dfrac{\mathrm{h} _{1}}{\mu _{1}}\right):\left(\dfrac{\mathrm{h} _{2}}{\mu _{2}}\right):\left(\dfrac{\mathrm{h} _{3}}{\mu _{3}}\right)$

(iii) $\left(\mathrm{h} _{2}^{2} \mathrm{~h} _{3}^{2}\right):\left(\mathrm{h} _{3}^{2} \mathrm{~h} _{1}^{2}\right):\left(\mathrm{h} _{1}^{2} \mathrm{~h} _{2}^{2}\right)$

(iv) $\left(\mu _{2}^{2} \mu _{3}^{2}\right):\left(\mu _{3}^{2} \mu _{1}^{2}\right):\left(\mu _{1}^{2} \mu _{2}^{2}\right)$

Show Answer

Solution:

We have

Apparent depth $=\dfrac{\text { Real depth }}{\mu}$

$$ =\dfrac{\mathrm{k}}{\mu^{2}} \quad\left(\because \text { Real depth } \alpha \dfrac{1}{\mu} \text { (as given) }\right) $$

$\therefore$ Ratio of apparent depths is

$$ \left(\dfrac{1}{\mu _{1}^{2}}: \dfrac{1}{\mu _{2}^{2}}: \dfrac{1}{\mu _{3}^{2}}\right) $$

or

$$ \left(\mu _{2}^{2} \mu _{3}^{2}: \mu _{3}^{2} \mu _{1}^{2}: \mu _{1}^{2} \mu _{2}\right) $$

Example:

A mark, put on a sheet of paper (kept on a table) is viewed from a point at a height $\mathrm{H}$ above the table. A glass slab (of refractive index $\mu$ and thickness $t$ ) is now put on the mark and the mark is again viewed from the same point. The depth of its image now, below the point of view, would be (nearly)

(i) $\left[\mathrm{H}+\mathrm{t}\left(1-\dfrac{1}{\mu}\right)\right]$

(ii) $\left(\mathrm{H}-\dfrac{\mathrm{t}}{\mathrm{\mu}}\right)$

(iii) $\left[\mathrm{H}-\mathrm{t}\left(1-\dfrac{1}{\mu}\right)\right]$

(iv) $\left(\mathrm{H}+\dfrac{\mathrm{t}}{\mu}\right)$

Show Answer

Solution:

The apparent depth of the glass slab would be $t^{\prime}$ where $\left(\dfrac{t}{t^{\prime}}\right) \cong \mu$. Hence $t^{\prime} \simeq \dfrac{t}{\mu}$. The mark, therefore, appears to be at a point $\left(t-\dfrac{t}{\mu}\right)$ above the table. Hence the depth of the image, with respect to the ‘point of view’, is $H-\left(t-\dfrac{t}{\mu}\right)=\left[H-t\left(1-\dfrac{1}{\mu}\right)\right]$.

EXPERIMENT-18

Characterstic curves of a pn junction diode in forward bias and reverse bias.

Introduction

A p-n junction is a diode which permits easy flow of currect in only one direction and offers high resistantce to the flow of current in the opposite direction. Fig. (a) shows a p-n junction in forward bias and Fig. (b) shows $\mathrm{p}-\mathrm{n}$ junction in reverse bias

The circuit diagram, for studying IV characterstics of a p-n junction diode, is shown below.

Forward Bias

The external voltage is applied to the diode in such a polarity that the positive terminal is connected to $\mathrm{p}-$ type end of the diode and the negative terminal is connected to n-type end of the diode. There occurs a sudden increases in the current flowing through the diode. We have, here;

1. Knee Voltage: It is the voltage across the diode in the forward bias configuration at which the current flowing through the diode starts increasing rapidly.

2. Breakdown Voltage: It is the voltage across the diode, in reverse biased configuration, at which the reverse current suddenly rises; the $\mathrm{p}-\mathrm{n}$ junction is then said to have ‘broken down’.

3. Avalanche Breakdown: In a normal $p-n$ junction diode, a large reverse voltage results in a large electric field due to which covalent bonds break, thereby giving rise to sudden increase in reverse current. The charge carriers, thus created, collide with other ions. This process is multiplicative and current increases like an avalanche. It is, therefore, referred to as ‘avalanche breakdown’.

The relation between current (I) flowing through the diode (measured by milliammeter), the voltage drop (V) across the diode (measured by connecting either a voltmeter or a multimeter), is given by.

$$ \mathrm{I}=\mathrm{I} _{\mathrm{S}}\left\{\exp \left(\dfrac{\mathrm{eV}}{\eta \mathrm{KT}}\right)-1\right\} $$

Here, Is $=$ reverse saturation current

$\mathrm{e}=$ charge of electron

$\eta=a$ constant whose value is 1 for Ge diode and (-2) for silicon diode.

$\mathrm{k}=$ Boltzman constant.

Therefore, in forward bias,

1. When external voltage is zero, then no current flows through the diode.

2. As voltage increases, potential barrier of the $\mathrm{p}-\mathrm{n}$ junction starts reducing. At a voltage $(0.3 \mathrm{~V}$ for $\mathrm{Ge}$ and $+0.7 \mathrm{~V}$ for $\mathrm{Si}$ ), the potential barrier is negligible and current starts increasing exponentially with voltage.

In this region, $\mathrm{I} \simeq \mathrm{I} _{\mathrm{S}} \exp \left(\dfrac{\mathrm{eV}}{\eta \mathrm{KT}}\right)$

Reverse Bias

The external voltage is applied to the diode such that, the positive terminal of the voltage is connected to ntype end of the diode and negative terminal is connected to $p$-type end of the diode. In this configuration, junction resistance is very large and the potential barrier of the junction is also large. Very small current flows through the diode due to the minority carriers.

If reverse voltage is increased, the kinetic energy of minority carriers increases; they may then remove electrons from the semiconductor atoms. This stage is called as breakdown of the junction; the voltage, at which it occurs, is called as the breakdown voltage.

EXPERIMENT-19

Characterstics curves of a Zener Diode and finding the Reverse Breakdown Voltage.

CONCEPT MAP

Introduction

Zener diode is a diode having heavily doped $\mathrm{p}$ and $\mathrm{n}$ regions of the $\mathrm{p}-\mathrm{n}$ junction diode. The Zener diode is specially doped / designed in a manner, so that in the breakdown region, it acts as a constant voltage source or as a voltage regulator.

The I-V characterstics curve of a Zener diode in forward bias configuration is similar to that of a $p-n$ junction diode. In the reverse bias, a Zener diode has a sharper ‘breakdown voltage’.

The breakdown voltage is sometimes called as Zener voltage; the corresponding current is called as ‘knee current’.

The breakdown voltage depends on the extent of doping. It doping is high, the depletion layer is very thin; as a result, the breakdown occurs even at a lower reverse voltage.

In order to determine the reverse breakdown voltage of a zener diode the following circuit is made.

The input voltage $\mathrm{V} _{\mathrm{i}}$ is slowly varied. A small current starts flowing through the Zener, but since Zener diode is reverse biased it offers high resistance to the current. Therefore, all the voltage drop occurs across the output terminals.

When $\mathrm{V} _{\mathrm{i}}>\mathrm{V} _{\mathrm{z}}$ (Zener voltage), the Zener diode will offer low resistance to current flow. Consequentaly, voltage drop now occurs across $\mathrm{R} _{\mathrm{s}}$. The voltage drop across $\mathrm{R} _{\mathrm{L}}$ remains constant at $\mathrm{V} _{\mathrm{z}}$. The Zener, in this range of input voltage, acts as a voltage regulator. The characterstic curve has the form shown.

EXPERIMENT–20

Characterstic curves of a transistor and finding current gain and voltage gain.

CONCEPT MAP

Introduction

We known that the transistor is a three terminal device that is used for a wide variety of practical applications.

The simple junction transistor has two well known forms: the $\mathrm{n} \mathrm{p} \mathrm{n}$ and the $\mathrm{p} \mathrm{p}$ transistor. We shall understand the basic features and characterstics of the transistor through a discussion on the pnp transistor.

A pnp transistor is normally biased in the manner shown here.

For a n p n transistor, all the polarities will change.

In normal transistor action, the emitter-base junction is forward biased and collector-base junction is reverse biased. Due to forward bias, the holes in the p-type emitter flow towards the n-types base. This flow of holes constitutes the emitter current $\left(\mathrm{I} _{\mathrm{E}}\right)$. These holes tend to recombine with the electrons present in the base. But since the base is only lightly doped, only few holes are able to recombine. The remaining holes cross into the p-type collector and make up the collector current.

Therefore, $\mathrm{I} _{\mathrm{E}}=\mathrm{I} _{\mathrm{B}}+\mathrm{I} _{\mathrm{C}}$

Some “Facts” about the Transistor

1. A transistor has 3 regions - emitter, base and collector. The base is the thinnest and is only lightly doped. Due to this, it passes most of the charge carriers from emitter to collector.

2. Emitter is heavily doped becuase it has to inject charge carriers into the base.

3. Collector region is the largest and is moderately doped.

Circuit diagram for $\mathrm{CE}$ transistor characterstics.

Input Characterstics

It is a curve between base curcuit $\mathrm{I} _{\mathrm{B}}$, and base emitter voltage $\mathrm{V} _{\mathrm{BE}}$, at constant collector emitter voltage, $\mathrm{V} _{\mathrm{CE}}$. A sample $\mathrm{I}-\mathrm{V}$ graph is plotted here. Notice,

1. It resembles the IV characterstics curve of a forward biased diode. It is so because the emitter-base junction is forward biased and acts like a diode.

2. Initially base current is very small. It increases sharply once the input voltage ( or $\mathrm{V} _{\mathrm{BE}}$ ) exceeds the barrier potential of

$\sim 0.7 \mathrm{~V}$.

3. If the (reverse) collector voltage is increased, then, for the same $\mathrm{V} _{\mathrm{BE}}$, the current decreases. The reason is as follows. The reverse biased collector attracts the majority charge carriers; therefore, the extent of recombination in the base decreases. Thus base current also decreases.

Output Characterstics

It is a curve between base curcuit $\mathrm{I} _{\mathrm{C}}$, and collector emitter voltage $\mathrm{V} _{\mathrm{CE}}$, at a constant base current, $\mathrm{I} _{\mathrm{B}}$. A simple IV graph is plotted here.

It should be noticed:

1. Initially collector current increases with base current. But after about $1 \mathrm{~V}$ (value of $\mathrm{V} _{\mathrm{CE}}$ ), collector current bcomes (almost) constant and independent of $\mathrm{V} _{\mathrm{CE}}$. This voltage (of about IV) is called as the knee voltage.

The small slope, in the IV curve, beyond knee voltage is because of the collector depletion layer which gets wide enough (due to reverse collector setting), to capture majority charge carriers.

2. Current gain: It is ratio of change in collector current to the change in base current.

$$ \beta=\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{B}}} $$

Generally value of $\beta$ lies between 20 to 500 . This is because of the fact, that less than $5 \%$ of emitter current flows as base current. The rest of the charge carriers move to collector and constitute the collector current.

In common base configuration, current gain is defined as

$$ \alpha=\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{E}}} \text { at constant } \mathrm{V} _{\mathrm{CB}} $$

Value of $\alpha$ is less than unity because a part of the emitter current flows as base current; hence $\mathrm{I} _{\mathrm{C}}$ is necessarily smaller than $\mathrm{I} _{\mathrm{E}}$.

3. Relation between $\alpha$ and $\beta$ :

$$ \beta=\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{B}}}=\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{E}}-\Delta \mathrm{I} _{\mathrm{C}}}=\dfrac{\Delta \mathrm{I} _{\mathrm{C}} / \Delta \mathrm{I} _{\mathrm{E}}}{\dfrac{\Delta \mathrm{I} _{\mathrm{E}}}{\Delta \mathrm{I} _{\mathrm{E}}}-\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{E}}}}=\dfrac{\alpha}{(1-\alpha)} $$

EXPERIMENT-21

Identification of diode, LED, transistor, IC, resistor, capacitor from a rised collection of such items.

Introduction

We know that the diode, LED, transistor, IC, resistors and capactiors are widely used in different electronic and electric circuits. It is important, therefore, to get familiar with their basic characterstics and the simple ways to identify them, from a mixed collection of such items.

We now talk about each of these devices in some detail.

1. Resistor

(i) A resistor is a small electric/electronic device that resists the flow of electric current. It obeys Ohm’s law which states that the current I flowing through a resistor (having resistance $\mathrm{R}$ ) is directly proportional to the voltage $\mathrm{V}$ across $\mathrm{R}$.

I $\alpha \mathrm{V} \Rightarrow \mathrm{IR}=\mathrm{V} \rightarrow$ Ohm’s law

(ii) The commanly used resistor in practical laboratories are the (two terminal) colour coded carbon resistors. They generally have four colours stips. The first three colour signify the the value of resistor the fourth colour implies the tolerance rating. The colour and their respective numbering is as follows:

Golden $= \pm 5 \%$ tolerance

Silver $= \pm 10 \%$ tolerance

B Black 0

B Brown 1

R Red 2

$\mathrm{O}$	Orange	3
Y	Yellow	4
G	Green	5
B	Blue	6
V	Violet	7
G	Grey	8
W	White	9

If the four folour are: yellow, red, red golden, it implies that the resistance

$$ =42 \times 10^2 \pm 5 \% \Omega=(4200 \pm 210) \Omega $$

(iii) Resistor is a passive device its the two terminals are not categorized as cathode / anode or as $(+\mathrm{ve}) /-\mathrm{ve})$.

(iv) Place the multimeter knob on the dial marked will the symbol ’ $\Omega$ ‘. The value of resistor can be directly read on the display by placing each of the two probes of multimeter on the two terminals of the resistance.

2. Diode and LED

(i) Diode and LED are semiconductor devices which consist of two semi conductor layers forming a p-n junction. LED (light emitting diode) is a special type of diode which has the capacity of emitting light.

(ii) The diode and the LED, both have two terminals and one of them is the (p-type) positive anode terminal. The other is the (n-type) negative cathode terminal.

The lengths of wires, joined to the positive and negative terminal, are different.

LED

(a)

• The longer leg is the $(+\mathrm{ve})$ anode terminal.

• The shorter leg is the ( $-\mathrm{ve})$ cathode terminal.

[Note: This is not the best identification criteria because some manufacturers offer the polarity in the reverse way also].

(b) If the lengths of the two terminals is same, then look inside the plastic casing. The flat-flag type end correponds to the cathode ( $-\mathrm{ve})$.

(c) There are no coloured striped markings in the LED.

(d) Generally, there are no number codes on an LED.

(e) Connect positive probe of the multimeter to the anode of LED and also place the negative (common) probe of multimeter on the cathode of LED.

Put the knob of the multimeter on the ‘diode’ symbol. You will see that LED will light up in this forward bias configuration. The display will show a value $\sim 1.7 \mathrm{~V}$ (depending on the colour of LED). If the probes are connected with opposite polarity, the LED will not glow and display of multimeter wil show 1. (This implies ‘overload’).

Diode

The anode and cathode of diode can be identified through the marking of a strip. Towards one end of the diode, a shaded band type marking is made. The signifies the cathode of the diode. Generally, normal diodes are black coloured with a grey-silver band type strip. [Remember: In a resistor, we have four coloured strips, but in a diode, we have only one strip!!]. Diode have low resistance to flow of current in one direction and a very high resistance in the other direction.

The diodes are always numbered. The numbering code, however, varies from one manufacturer to another. For example in American system, the code is like IN4007 which implies that it is a normal, diode, having tolerance of $1000 \mathrm{~V}$ and forward voltage drop of $0.6 \mathrm{~V}$ and an average forward current of 1A.

By using multimeter, the diode can be identified as follows:

Connect multimeter probes to the diode terminals as per its forward biased configuration. Put the multimeter knob on ‘diode’ symbol. The display will show $\sim 0.587 \mathrm{~V}$. This is the forward voltage drop between the two leads. In the reverse polarity connection, the display will read 1 . This implies infinite (very large) resistance of the diode in its reverse biased configuration.

3. Capacitor

Capacitor is also a two terminal device, but (as compared to resistance and diode) it is generally not colour coded. Most capacitor have capacitance value written on them.

(a) This is typical example of non-electrolytic (casing) capacitor. There is no anode / cathode

$$ \begin{aligned} 102 & \Rightarrow 10 \times 10^{2} \mathrm{P}^{\mathrm{F}}=10 \times 10^{2} \times 10^{-12} \mathrm{~F} \\ & =1 \mathrm{nF} \end{aligned} $$

(b)

• Electrolytic cylindrical aluminimum casing.

4.7 $\mu \mathrm{F}$ is directly written on the cap.

• The two terminals are of different lengths.

• Longer leg is +ve (anode).

• Shorter leg is-ve (cathode).

• Sometimes ‘-’ minus sign is also drawn on the cap.

4. Transistor

Transistor is a three terminal device; it can be easily identified from a collection of items ( $\mathrm{R}, \mathrm{C}$, diode, LED). We will talk more about the transistor in the next experiment.

5. IC

Integrated circuit (IC) is one in which components such as diodes, transistors, capacitors and resistors etc. are embeded on a small semiconductor chip. The size of IC is so small that one will require a microscope to see the connection between these components. The ICs are protected from external disturbances by encapsulating them in plastic / metal/ ceramic packages. We generally have the IC package shown here.

It is a flat pack container having equal number of pins on either side. The total number of pins can be $8,10,12,14$ etc. depending on the function of IC.

Example:

The colour coding of a resistor is brown, black, red, golden.

(i) What will be the most likely value displayed when the multimeter knob is at ’ $2 \mathrm{~K}$ ‘?

(a) 1.003

(b) 1.00

(c) 01.0

(d) 0.001

(ii) What will be the most likely value displayed when knob is at ’ $20 \mathrm{~K}$ ‘?

(a) 1.003

(b) 1.00

(c) 01.0

(d) 0.001

(iii) What will be the most likely value displayed when knob is at ’ $200 \mathrm{~K}$ ‘?

(a) 1.003

(b) 1.00

(c) 01.0

(d) 0.001

(iv) What will be the most likely value displayed when knob is at ’ $2 \mathrm{M}$ ‘?

(a) 1.003

(b) 1.00

(c) 01.0

(d) 0.001

Solution:

(i) (a)

(ii) (b)

(iii) c

(iv) (d)

Reason: Brown black red $\Rightarrow 10 \times 10^{2} \Omega=1000 \Omega$

Golden $\Rightarrow$ tolerance as $5 \%$ of $1000=50 \Omega$

Value of resistor lies between $950-1050 \Omega$

For best precision, knob should be placed at the next higher position i.e. $2 \mathrm{~K}$,

where $1.003 \rightarrow 1.003 \mathrm{~K} \Omega=1003 \Omega$

Precision in measurement decreases if the knob is placed at other higher markings.

EXPERIMENT-22

Use multimeter to identify: base of a transistor; distinguish between npn and pnp transistor; see unidirectional flow of current in case of a diode and a LED.

CONCEPT MAP

Introduction

We know that the multimeter is an integral and essential part of an electronics or physics laboratory. It is put to many uses. Here we will try to understand its uses with respet to a transistor and a diode.

The basic details of the use of a multimeter for doing the indicated identifications, in case of a transistor and a diode/LED, are outlined below. We first do a brief recapitulation of the basic features of a multimeter.

The Multimeter

Multimeter basically has 3 important parts:

(a) Display Section: Here generally 4 digits value, and +/sign of the quantity being measured, is displayed.

(b) Selection Knobs: It allows the user to set the multimeter to read different quantities such as current (ac/dc), voltage (ac/dc), resistance, capacitance etc.

(c) Two probes which are inserted in the ports present at the bottom. One of the probe is connected to COM (ground) port; the other is connected to either of the ports depending on the quantity to be measured.

Identification of Base of Transistor

Physically

(i) In some transistors, there is a notch. The leg / pin, next / nearest to the notch, is emitter. Moving ahead in clockwise direction (pins towards the user) gives base and the collector.

(ii) Some transistors have a black flat top which is rounded on one end. Keep flat surface towards you and the pins pointing upwards. Emitter, Base and collector will be as shown in the diagram.

Using a Multimeter

(i) Transistor are either $\mathrm{n}-\mathrm{p}-\mathrm{n}$ or $\mathrm{p}-\mathrm{n}-\mathrm{p}$, i.e. basically two diodes placed side by side. In $\mathrm{n}-\mathrm{p}-\mathrm{n}$

Put -ve probe of multimeter to one of the pins. Put selection knob on the ‘diode’ symbol. Put +ve probe to other 2 pins. A value $0.5-0.7$ will be displayed in only one configuration. This happens when $-v e$ probe is connected to $n$ and the $+v e$ probe is connected to $p$. Repeat the process. You will find that, whenever +ve probe is kept at pin *, then $0.5-0.7$ is always displayed. The pin * will be BASE.

Only in $\mathrm{p}-\mathrm{n}-\mathrm{p}$ transistor

Whenever -ve probe is at pin *, then $0.5-0.7$ will always be displayed. Pin * will be BASE.

Distinguish between npn and pnp Transistor

In the previous article, we identified the base of transistor. Once base is identified, the type of transistor, $\mathrm{npn} / \mathrm{pn} \mathrm{p}$ is automatically determined.

There is one more method.

Put knob on ’ $h _{\mathrm{fe}}$ ‘. Multimeters have grooves such that

Put E,B, C of transistor in (1). If it is npn transistor, then it will show $h _{f e}$ value in display. $h _{f e}$ is forward current gain of transistor. If transistor is pnp type, then it will show a reading / value only when placed in (2).

Unidirectional Flow of Current through Diode / LED

We have seen in previous experiment, that when knob of multimeter is placed on ‘diode’ symbol, then diode shows $0.5-0.7$ on display panel. LED glows up when forward biased. This establishes the unidiretional flow of current through diode and LED.

PROBLEMS FOR PRACTICE

1. The main scale, of a given vernier scale set up, has been gradiented to read up to $(1 / 2)^{0}$. There are 30 division on its vernier scale that coincide with 29 division of its main scale.

The zero error of this vernier equals (-2) times its least count. The readings, in a given case, are as follows:

Main scale reading $=63 \cdot 5^{\circ}$. Coincident vernier scale division $=17^{\text {th }}$.

The corrected value of this reading is

(1) $63^{\circ} 49^{\prime}$

(2) $63^{\circ} 45^{\prime}$

(3) $63^{0} 15$,

(4) $63^{0} 11$ '

Show Answer

Answer: (1)

2. In a given vernier callipers, $n$ division, on the vernier scale coincide with ( $\mathrm{n}-1$ ) divisons of the main scale. The main scale has been graduated to read up to $(x)$ units. If the zero error is $(+p)$ times th least count, the reading when the

(i) Main scale reads (m) units

(ii) The $\left(\mathrm{k}^{\text {th }}\right)$ division on the vernier scale coincides with a main scale division, is

(1) $\left[\mathrm{m}+\dfrac{x}{\mathrm{n}}(\mathrm{k}+\mathrm{p})\right]$

(2) $\left[\mathrm{m}+\dfrac{x}{\mathrm{n}}(\mathrm{k}-\mathrm{p})\right]$

(3) $\left[\mathrm{m}+\left(\dfrac{x}{\mathrm{n}-1}\right)(\mathrm{k}-\mathrm{p})\right]$

(4) $\left[\mathrm{m}+\left(\dfrac{x}{\mathrm{n}-1}\right)(\mathrm{k}+\mathrm{p})\right]$

Show Answer

Answer: (2)

3. A given screw gauge has a least count of $0.002 \mathrm{~cm}$ and there are 50 divisons on its circular scale. When the jaws, of this screw guage, are just in contact with each other, the $7^{\text {th }}$ division of the circular scale lies along its ‘reference line’.

When this scrw gauge, is used to measure the thickness of a thin mica sheet, the main scale reads 0.7 $\mathrm{cm}$ and the $21^{\text {st }}$ division, of its circular scale, coincides with the ‘reference line’.

The ‘pitch’ of this screw gauge, and the thickness of the mica sheet, are respectively, equal to

(1) $0.2 \mathrm{~cm}$ and $0.717 \mathrm{~cm}$

(2) $0.1 \mathrm{~cm}$ and $0.756 \mathrm{~cm}$

(3) $0.2 \mathrm{~cm}$ and $0.728 \mathrm{~cm}$

(4) $0.2 \mathrm{~cm}$ and $0.756 \mathrm{~cm}$

Show Answer

Answer: (3)

4. The pitch, of a given screw guage, equals (p) $\mathrm{cm}$ and its circular scale has $\mathrm{n}$ division on it. When the jaws of the screw gauage are just in contact with eath other, the ‘zero’, of the circular scale, lies along the reference line.

When a thin wire is just held between the jaws of the screw guage, the main scale reads $(\mathrm{m})$ and the $\mathrm{k}^{\text {th }}$ division, of the circular scale, lies along the ‘reference line’. The volume, of a length $\mathrm{L}$, of this wire, is

(1) $\dfrac{\pi \mathrm{L}}{4}\left[\left(\mathrm{~m}+\dfrac{\mathrm{p}}{\mathrm{n}} \cdot \mathrm{k}\right)^{2}\right]$

(2) $\pi \mathrm{L}\left[\left(\mathrm{m}+\dfrac{\mathrm{p}}{\mathrm{n}} \cdot \mathrm{k}\right)^{2}\right]$

(3) $\dfrac{\pi \mathrm{L}}{4}\left[(\mathrm{~m}-\mathrm{pnk})^{2}\right]$

(4) $\pi \mathrm{L}\left[(\mathrm{m}-\mathrm{pnk})^{2}\right]$

Show Answer

Answer: (1)

5. The energy, of the bob of a simple pendulum, decreases to $64 \%$ of its original value. The amplitude of oscillations must have decreased to

(1) $80 \%$ of its original value

(2) $75 \%$ of its original value

(3) $72 \%$ of its original value

(4) $64 \%$ of its original value

Show Answer

Answer: (1)

6. The graph showing the variation, of the square of the amplitude $\left(\mathrm{A}^{2}\right)$ of the oscillating bob of a simple pendulum, with time, is the graph labelled as graph

(1) $\mathrm{P}$

(2) $\mathrm{Q}$

(3) $\mathrm{R}$

(4) $\mathrm{S}$

Show Answer

Answer: (2)

7. The meter scale gets balanced in the set-up shown here. The mass, $x$, equals (nearly)

(1) $8.2 \mathrm{~g}$

(2) $8.8 \mathrm{~g}$

(3) $9.2 \mathrm{~g}$

(4) $9.8 \mathrm{~g}$

Show Answer

Answer: (4)

8. The mass of the meter scale that gets balanced, in the set-up shown, equals (nearly)

(1) $5 \mathrm{~g}$

(2) $10 \mathrm{~g}$

(3) $15 \mathrm{~g}$

(4) $20 \mathrm{~g}$

Show Answer

Answer: (4)

9. Two wires, made of materials $A$ and $B$ (having Young’s moduli $Y _{A}$ and $Y _{B}$ ) have unloaded lengths $\ell _{\mathrm{A}}$ and $\ell _{\mathrm{B}}$, respectively. They are joined and loaded as shown. If the diameter of each wire is $d$, the extensions, produced in them, are equal, respectively, to

(1) $\dfrac{4\left(\mathrm{M} _{1}+\mathrm{M} _{2}\right) \mathrm{g} \ell _{\mathrm{A}}}{\left(\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{A}}\right)}$ and $\left(\dfrac{4 \mathrm{M} _{2} \mathrm{~g} \ell _{\mathrm{B}}}{\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{B}}}\right)$

(2) $\dfrac{4\left(M _{1} g\right) \ell _{A}}{\pi d^{2} Y _{A}}$ and $\left(\dfrac{4 M _{2} g \ell _{B}}{\pi d^{2} Y _{B}}\right)$

(3) $\dfrac{4\left(\mathrm{M} _{1} \mathrm{~g}\right) \ell _{\mathrm{A}}}{\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{A}}}$ and $\dfrac{4 \mathrm{M} _{2} \mathrm{~g}\left(\ell _{\mathrm{A}}+\ell _{\mathrm{B}}\right)}{\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{B}}}$

(4) $\dfrac{4\left(\mathrm{M} _{1}+\mathrm{M} _{2}\right) \mathrm{g}\left(\ell _{\mathrm{A}}+\ell _{\mathrm{B}}\right)}{\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{A}}}$ and $\dfrac{4 \mathrm{M} _{2} \mathrm{~g}\left(\ell _{\mathrm{A}}+\ell _{\mathrm{B}}\right)}{\pi \mathrm{d}^{2} \mathrm{Y} _{\mathrm{B}}}$

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Answer: (1)

10. N idential hollow cylindrical columns, made from a metal of Young’s modulus $Y$, together support uniformly a total load of mass $\mathrm{Mkg}$. The inner and outer radii of each column, are $\mathrm{r}$ and $\mathrm{R}$, respectively. The compressional strain, developed in each column, equals

(1) $\dfrac{(\mathrm{Mg})}{\left(\mathrm{N} \pi \mathrm{YR}^{2}\right)}$

(2) $\dfrac{(\mathrm{NMg})}{\left(\pi \mathrm{YR}^{2}\right)}$

(3) $\dfrac{(\mathrm{NMg})}{\left(\pi \mathrm{Y}\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)\right)}$

(4) $\dfrac{(\mathrm{Mg})}{\left(\pi \mathrm{NY}\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)\right)}$

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Answer: (4)

11. Water, (assumed to ‘spread out’ on glass), rises to a height $\mathrm{h}$ (above the general outside level) in a (very small radius) glass capillary tube. The surface tension of water may be taken as $70 \mathrm{~N} / \mathrm{m}$.

If the same capillary tube were to be dipped in mercury (surface tension $\simeq 560 \mathrm{~N} / \mathrm{m}$, relative density $=13.35$, angle of contact with glass $=\left(\dfrac{\pi}{2}+\alpha\right)(\alpha$ is an acute angle $)$, the level of mercury, in the capillary tube, relative to the general outside level, would be

(1) at a height h’ where h’ $\simeq-1.67 \cos \alpha$

(2) at a depth $\mathrm{d} _{1}$ where $\mathrm{d} _{1} \simeq 0.6 \cos \alpha$

(3) at a depth $d _{2}$ where $d _{2} \simeq 0.6 \sin \alpha$

(4) at a height h" where $\mathrm{h}^{\prime \prime} \simeq-1.67 \sin \alpha$

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Answer: (3)

12. The difference, in the mean value of the readings of the travelling microscope, when its horizontal ‘cross wire’, is at the two positions, indicated here, equals $x$.

This capillary tube is dipped in a liquid of surface tension $T$ and density $\rho$. The angle of contact, of

the liquid with the walls of the capillary tube, is $\alpha$. If the tube has its axis inclined at an angle $\alpha$ to the vertical, the length of the liquid column (relative to the general outside level), in the capillary tube, would equal.

(1) $\dfrac{(4 \mathrm{~T} \cos \theta)}{(x \rho g \cos \alpha)}$

(2) $\dfrac{(4 \mathrm{~T} \cos \alpha)}{(x \rho \cos \theta)}$

(3) $\dfrac{(4 \mathrm{~T} \cos \alpha \cos \theta)}{(x \rho g)}$

(4) $\dfrac{(4 \mathrm{~T})}{(x \rho g \cos \alpha \cos \theta)}$

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Answer: (2)

13. An oil drop (density of oil $=\rho$ ), of radius $r$, falls vertically in the region between two plates, having a separation $d$ and a potential difference $V$ between them. The charge on the drop is (-q). Taking the density of air as $\sigma$ and the coefficient of viscosity of air as $\eta$, the terminal velocity acquired by the drop (assuming the electrical force on it. to be smaller than the gravitational force on it), would be

(1) $\left[\dfrac{9}{2} \dfrac{\mathrm{r}^{2}(\rho-\sigma) \mathrm{g}}{\eta}-\dfrac{\mathrm{qV}}{6 \pi \eta \mathrm{r}}\right]$

(2) $\left[\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}-\dfrac{q V}{6 \pi \eta r}\right]$

(3) $\left[\dfrac{2}{9} \dfrac{r^{2}(\rho-\sigma) g}{\eta}-\dfrac{q V}{6 \pi \eta r d}\right]$

(4) $\left[\dfrac{9}{2} \dfrac{\mathrm{r}^{2}(\rho-\sigma) \mathrm{g}}{\eta}-\dfrac{\mathrm{qV}}{6 \pi \eta \mathrm{rd}}\right]$

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Answer: (3)

14. Adrop of a liquid (of density $\rho$ ), having a radius $r$, falls through a viscous medium of density $\sigma$ and coefficient of viscosity $\eta$. It acquires a terminal velocity $\mathrm{v} _{1}$.

Another drop of another liquid (of density $\dfrac{\rho}{\sqrt{N}}$ ), has a volume equal to the volume of $\mathrm{N}$ drops (each of radius r) of the first liquid. It falls through a viscous medium (of density $\dfrac{\rho}{\sqrt{N}}$ ) and coefficient of viscosity $\mathrm{N} \eta$. It acquires a terminal velocity $\mathrm{v} _{2}$. The ratio $\left(\dfrac{\mathrm{v} _{2}}{\mathrm{v} _{1}}\right)$ equals.

(1) $\left(\mathrm{N}^{2 / 3}\right)$

(2) $\dfrac{1}{\left(\mathrm{~N}^{1 / 3}\right)}$

(3) $\left(\mathrm{N}^{1 / 3}\right)$

(4) $\dfrac{1}{\left(\mathrm{~N}^{5 / 6}\right)}$

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Answer: (4)

15. The (temperature) vs (time) graph, of a body (initially at a temperature $\theta _{1}$, and present in surroundings at a temperature $\theta _{0}$ ), would have the form corresponding to the graph labelled as graph

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

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Answer: (1)

16. Ahot body, initially at a temperature $\theta$, is present in surroundings that are at a temperature $\theta _{0}$. The times, taken by the hot body, to cool from:

(i) $\left[\theta \text { to }\left(\dfrac{\theta}{2}\right)\right] \text { is }(\Delta t)_1$

(ii) $\left[\left(\dfrac{\theta}{2}\right)\right.$ to $\left.\left(\dfrac{\theta}{4}\right)\right]$ is $(\Delta \mathrm{t}) _{2}$

We can then say that

(1) $(\Delta \mathrm{t}) _{1}=(\Delta \mathrm{t}) _{2}$

(2) $(\Delta \mathrm{t}) _{1}=\dfrac{1}{2}(\Delta \mathrm{t}) _{2}$

(3) $(\Delta \mathrm{t}) _{1}<(\Delta \mathrm{t}) _{2}$

(4) $(\Delta t) _{1}>(\Delta t) _{2}$

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Answer: (3)

17. A student does the experiment on finding the speed of sound using a resonance tube. He uses a resonance tube of diameter $\mathrm{D}$ and a tuning fork of frequency $\mathrm{N}$. If the velocity of sound, at the room temperature is $\mathrm{V}$, the first and second resonance lengths would equal, respectively.

(1) $\dfrac{\mathrm{V}}{4 \mathrm{~N}}$ and $\dfrac{3 \mathrm{~V}}{4 \mathrm{~N}}$

(2) $\left(\dfrac{\mathrm{V}}{4 \mathrm{~N}}-0.3 \mathrm{D}\right)$ and $\left(\dfrac{3 \mathrm{~V}}{4 \mathrm{~N}}-0.3 \mathrm{D}\right)$

(3) $\left(\dfrac{\mathrm{V}}{4 \mathrm{~N}}+0.3 \mathrm{D}\right)$ and $\left(\dfrac{3 \mathrm{~V}}{4 \mathrm{~N}}+0.3 \mathrm{D}\right)$

(4) $\left(\dfrac{\mathrm{V}}{4 \mathrm{~N}}+0.3 \mathrm{D}\right)$ and $\left(\dfrac{3 \mathrm{~V}}{4 \mathrm{~N}}+0.9 \mathrm{D}\right)$

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Answer: (2)

18. The speed of sound in air, at $0^{\circ} \mathrm{C}$ equals $\mathrm{V} _{0}$. A student does the experiment, on finding the speed of sound, at room temperature $\left(=\theta^{0} \mathrm{C}\right)$ using a resonance tube. The resonance tube has a diameter $\mathrm{D}$ and the tuning fork used has a frequency $\mathrm{N}$. The difference, between the second and first resonance lengths, would then equal

(1) $\dfrac{\mathrm{V} _{0}}{2 \mathrm{~N}}\left(1+\dfrac{\theta}{273}\right)^{1 / 2}$

(2) $\dfrac{\mathrm{V} _{0}}{\mathrm{~N}}\left(1+\dfrac{\theta}{273}\right)^{-1 / 2}$

(3) $\dfrac{\mathrm{V} _{0}}{2 \mathrm{~N}}\left(1+\dfrac{\theta}{273}\right)^{-1 / 2}$

(4) $\dfrac{\mathrm{V} _{0}}{\mathrm{~N}}\left(1+\dfrac{\theta}{273}\right)^{1 / 2}$

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Answer: (1)

19. A given solid, of mass $\mathrm{M}$, is heated to a temperature $\theta^{\circ} \mathrm{C}$. It is put in contact with a large block of ice. It is known that a unit mass of ice needs $L$ units of heat to just melt down. If the mass of ice, that just melts down, is $m$, the specific heat capacity, of the material of the solid, is

(1) $\dfrac{\mathrm{mL}}{\mathrm{M} \theta}$

(2) $\dfrac{\mathrm{M} \theta}{\mathrm{mL}}$

(3) $\dfrac{\mathrm{ML}}{\mathrm{m} \theta}$

(4) $\dfrac{\mathrm{m} \theta}{\mathrm{ML}}$

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Answer: (1)

20. The quantity of heat needed to bring a rise $\theta$ in the temperature of a

(i) mass $\mathrm{m} _{1}$, of a liquid of specific heat capacity $\mathrm{s} _{1}$, is $\mathrm{H} _{1}$.

(ii) mass $\mathrm{m} _{2}$, of a solid of specific heat capacity $\mathrm{s} _{2}$, is $\mathrm{H} _{2}$.

If the quantity of heat needed to bring the same rise $\theta$ in the temperature of a mass $\left(m _{1}+m _{2}\right)$ of a substance of specific heat capacity $\left(\mathrm{s} _{1}+\mathrm{s} _{2}\right)$, is $\mathrm{H}$, the difference $\left[\mathrm{H}-\mathrm{H} _{1}+\mathrm{H} _{2}\right)$, equals

(1) zero

(2) $\left(\mathrm{m} _{1} \mathrm{~s} _{1}+\mathrm{m} _{2} \mathrm{~s} _{2}\right) \theta$

(3) $\left(\mathrm{m} _{1} \mathrm{~s} _{2}+\mathrm{m} _{2} \mathrm{~s} _{1}\right) \theta$

(4) $\left(\mathrm{m} _{1} \mathrm{~s} _{2}-\mathrm{m} _{2} \mathrm{~s} _{1}\right) \theta$

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Answer: (3)

21. An equilateral prism is made from a material of refractive index $\mu$. The angles of minimum deviation, for this prism, when kept in (i) air (ii) a transparent medium of refractive and $\mu^{\prime}$, are found to be $D$ and $D^{\prime}$, respetively. We would then have

(1) $\dfrac{\sin \left(30^{\circ}+\dfrac{D^{\prime}}{2}\right)}{\sin \left(30^{\circ}+\dfrac{D}{2}\right)}=\mu^{\prime}$

(2) $\dfrac{\sin \left(30^{\circ}+\dfrac{D^{\prime}}{2}\right)}{\sin \left(30^{\circ}+\dfrac{D}{2}\right)}=\dfrac{\mu}{\mu^{\prime}}$

(3) $\dfrac{\sin \left(30^{\circ}+\dfrac{D}{2}\right)}{\sin \left(30^{\circ}+\dfrac{D^{\prime}}{2}\right)}=\dfrac{\mu}{\mu^{\prime}}$

(4) $\dfrac{\sin \left(30^{\circ}+\dfrac{D^{\prime}}{2}\right)}{\sin \left(30^{\circ}+\dfrac{D}{2}\right)}=\dfrac{1}{\mu^{\prime}}$

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Answer: (4)

22. A ray of light, incident on one face of an equilateral prism (made from a material of refractive index $\mu$ ) undergoes total internal reflection, when its refracted ray falls on the other face of the prism. The angle of incidence, of this ray, equals.

(1) $\sin ^{-1}\left[\mu\left(60^{\circ}-\sin ^{-1} \dfrac{1}{\mu}\right)\right]$

(2) $\sin ^{-1}\left[\dfrac{\left(60^{0}-\sin ^{-1} \dfrac{1}{\mu}\right)}{\mu}\right]$

(3) $\cos ^{-1}\left[\left(60^{0}-\sin ^{-1} \dfrac{1}{\mu}\right)\right]$

(4) $\cos ^{-1}\left[\mu\left(60^{0}-\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

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Answer: (1)

23. The forward and reverse currents, flowing through a diode, are due respectively to

(1) Majority charge carriers for both the currents.

(2) Minority charge carriers for both the currents.

(3) Majority charge carriers and minority charge carriers.

(4) Minority charge carriers and majority charge carriers.

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Answer: (3)

24. The knee voltage, for Ge and Si diode are (respectively)

(1) $0.3 \mathrm{eV}, 0.7 \mathrm{eV}$

(2) $0.3 \mathrm{~V}, 0.7 \mathrm{~V}$

(3) $0.7 \mathrm{eV}, 0.3 \mathrm{eV}$

(4) $0.7 \mathrm{~V}, 0.3 \mathrm{~V}$

Show Answer

Answer: (2)

25. Zener diode, having a Zener voltage $\mathrm{V} _{\mathrm{Z}}$, is said to be in ’ $\mathrm{OFF}$ ’ state when, reverse voltage $\mathrm{V} _{\mathrm{R}}$ is such that

(1) $\mathrm{V} _{\mathrm{R}}>\mathrm{V} _{\mathrm{Z}}$

(2) $\mathrm{V} _{\mathrm{R}}=\mathrm{V} _{\mathrm{Z}}$

(3) $\mathrm{V} _{\mathrm{R}}>0$

(4) $\mathrm{V} _{\mathrm{R}}<\mathrm{V} _{\mathrm{Z}}$

Show Answer

Answer: (4)

26. A zener diode is generally used as,

(1) p-n junction diode

(2) Full wave rectifier

(3) Half wave rectifier

(4) Voltage regulator

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Answer: (4)

27. A characterstics feature of IV graph of a Zener diode is

(1) Gradual reverse breakdown

(2) Gradual forward breakdown

(3) Sudden reverse breakdown

(4) Sudden forward breakdown

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Answer: (3)

28. In a CE transistor (in active mode), the emitter-base junction has a resistance ( $r$ ) while the collectorbase junction has a resistanc $\mathrm{R}$. We can say that $\mathrm{r}$ and $\mathrm{R}$, are, respectively, a

(1) low resistance, low resistance

(2) high resistance, high resistance

(3) low resistance, high resistance

(4) high resistance, low resistance

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Answer: (3)

29. The input characterstics curves of a transistor, are curves between

(1) Input current and output voltage for fixed input voltage.

(2) Input current and input voltage for fixed output voltage.

(3) Output current and output voltage for fixed input voltage.

(4) Output current and input voltage for fixed output voltage.

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Answer: (2)

30. Two multimeters (A and B) require a total markings of $20 \mathrm{~mA}$ and $60 \mathrm{~mA}$, respectively, to give full scale deflection. It implies that,

(1) A will measure $10 \mathrm{~mA}$ current more accurately

(2) B will measure $10 \mathrm{~mA}$ current more accurately

(3) A and $\mathrm{B}$ will give equally accurate value of $10 \mathrm{~mA}$ current.

(4) B has a better ’least count’ than A.

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Answer: (1)

Question Bank

Key Learning Points

Experiment-1: Vernier Callipers

1. The vernier callipers has a vernier scale that slides on its main scale.

2. The least count of a vernier equals the difference between the values of its 1 main scale division and 1 vernier scale division. It is usually calculated by using the ‘formula’

$$ \text { Least count }=\dfrac{\text { Value of one main scale division }}{\text { No.of divisions on the vernier scale }} $$

3. The read a vernier, we read / observe

(i) the main scale reading $(\mathrm{m})$ just before the zero of the vernier scale.

(ii) the number (n) of the vernier scale division the coincides with some division on the main scale.

The complete reading is then

$$ \mathrm{m}+\mathrm{n}(\text { least count }) $$

4. A vernier callipers is said to have a ‘zero error’ if the zeroes, of its main scale and its vernier scale, do not coincide when the jaws of the vernier just touch each other.

5. The zero error is positive / negative if the zero of the vernier scale lies to the right / left of the zero of the main scale. (when the jaws of the vernier just touch other).

6. The ‘zero error’ is always subtracted algebraically, from the observed reading, to get the corrected reading.

Experiment-2: Screw Gauge

7. Ascrew gauge has an accurately threaded screw, having a closely fitting nut whose threadings match that of the screw.

8. The screw, when rotated, has a simultaneous rotatory motion along a circular path and a translatory motion along a straight path.

9. The linear distance, moved by the screw in one complete rotation, is known as its ‘pitch’.

10. The least count, of a screw gauge, is obtained by dividing its pitch, by the number of divisions on its circular scale.

11. To read a screw gauge, the ‘object’/dimension,to be measured, is just held between the jaws of the screw gauge. One then notes its main scale reading (say ’ $m$ ‘) and also notes the number (say ’ $k$ ‘), of the circular scale division, that just lies along the ‘reference line’. The reading of the screw gauge is then $(\mathrm{m}+\mathrm{k} \times \ell . \mathrm{c})$.

12. The screw gauge is said to have a ‘zero error’ if the zero of its circular scale, lies above / below the ‘reference line’, when the jaws of the screw gauge are just in contact with each other.

13. The magnitude of the ‘zero error’ is obtained by multiplying the ’least count’ with the number of divisons of the circular scale, by which its zero is above / below the reference line.

14. The ‘zero error’ is negative if the ‘zero’, of the circular scale (when the jaws of the screw gauge are just in contact with each other) lies above the ‘reference line’. It is positive if this lies below the reference line.

15. The ‘zero error’ is always subtracted algebraically from the observed readings.

16. A screw gauge, that has been in use for long (or has a manufacturing defect), can develop some ‘play’ in the motion of its screw. The screw then may not advance even when it is being rotated. This error is known as ‘backlash error’.

17. The ‘backlash error’ can be avoided by rotating the screw in one direction only.

Experiment-3: Simple Pendulum

18. A simple pendulum is made by attaching a small heavy spherical metal bob to a good quality sewing (cotton) thread.

19. The oscillations of a simple pendulum are simple harmonic only if the amplitude of its oscillations is small.

20. The oscillating bob, of a simple pendulum, has both kinetic as well as potential energies.

21. The total energy of the oscillating bob of the pendulum, equals $\dfrac{1}{2} \mathrm{~mA}^{2} \omega^{2}$ where $\mathrm{m}=$ mass of the bob, $\mathrm{A}=$ amplitude of oscillation; $\omega\left(=\dfrac{2 \pi}{\text { Time Period }}\right)$ is the angular frequency of the oscillating bob.

22. When the bob is oscillating in a medium (like air), its oscillations are damped (harmonic) oscillations.

23. The amplitude of (damped) oscillations decays exponentially with time.

24. One can observe the variation of amplitude with time, and use it to draw a graph between the square of amplitude (a measure of the energy) of the oscillating bob, with time.

25. The above graph is expected to be an exponentially decaying curve.

Experiment-4: Principle of Moments

26. We define the magnitude of the moment of a force, about a given point (or axis), as the product of the force and the ‘force-arm’. Here the ‘force-arm’ equals the perpendicular distance between the line of action of the force and the axis (or point) about which rotation is taking place.

27. The vector relation giving the magnitude as well as the direction of the moment $(=\vec{\tau})$ of a force $\vec{F}$, producing rotation about some point (say $\mathrm{O}$ ) is.

$$ \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}} $$

Here $\overrightarrow{\mathrm{r}}$ is the position vector of the point of application of the force with respect to the point $\mathrm{O}$.

28. The principle of moments says that

“Whenever a body is in equilibrium under the action of a number of coplaner forces, the algebraic sum of the moments of all there forces, about any arbitrary point, is zero.

29. We take anticlockwise moments as +ve and clockwise moments as (-)ve.

30. We can use the principle of moments to find an unknown mass, by using a meter scale and a known mass.

31. We can also use the principle of moments to find the mass of the meter scale itself.

Experiment-5: Young’s Modulus

32. “The tendency, to regain the original conditions, after the removal of the deforming forces”. This characterstics of materials is known as elasticity.

33. The stress developed, in a deformed object, equals the internal restoring force, measured per unit area.

34. Strain equals the ‘change in dimension’ per unit dimension.

35. According to Hook’s law:

“For smalls values of strain the stress, developed in an object, is proportional to the strain produced in it”.

36. Within the elastic limit:

$$ \dfrac{\text { Stress }}{\text { Strain }}=a \text { constant } $$

37. $\mathrm{Y}=\dfrac{\text { Longitudinal stress }}{\text { Longitudinal strain }}=$ Young’s modulus

38. $\mathrm{K}=\dfrac{\text { Volume stress }}{\text { Volumestrain }}=$ Bulk modulus

39. $\eta=\dfrac{\text { Shearing stress }}{\text { Shearing strain }}=$ Modulus of rigidity

40. The Si unit, of all moduli of elasticity, equals $\text { newton/metre }{ }^2$.

41. $\sigma=\dfrac{\text { Lateral strain }}{\text { Longitudinal strain }}=$ Poisson’s ratio

42. Searle’s appraratus is used for experimental measurement of Young’s modulus of the material of a wire.

Experiment-6: Surface Tension

43. The forces, which the atoms / molecules of a material exert on one another, are known as interatomic / intermolecules forces.

44. Interatomic / intermolecular forces are short ranged altractive forces; however, these forces become repulsive in nature if the atoms / molecules come too close to each other.

45. Interatomic / intermolecular forces are known as cohesive forces if they exist between similar atoms / molecules. For different types of atoms / molecules, these forces are known as adhesive forces.

46. The property of a liquid surface, to be in a state of stress, is referred to as the phenomenon of surface tension.

47. The surface tension of a liquid is measured in unit of $\mathrm{N} / \mathrm{m}$. Its dimensions are $\left[\mathrm{MT}^{-2}\right]$.

48. The shape of the meniscus of a liquid, when put in a container, is used to define a term called the ‘angle of contact’.

49. The addtional potential energy, of the molecules of a liquid on the surface of the liquid, is referred to as their surface energy.

50. The surface energy per unit area of the liquid surface, equals the surface tension of the liquid.

51. There exists an ’excess of pressure’ on the concave side of the curved surface of a liquid.

52. The excess pressure, on the concave side of the curved surface

(i) equals ( $2 \mathrm{~T} / \mathrm{r})$, for a drop of the liquid.

(ii) equals ( $4 \mathrm{~T} / \mathrm{r})$, for a bubble blown out of the liquid.

Here $\mathrm{T}$ is the surface tension of the liquid and $\mathrm{r}$ is the rdius of curvature of its curved surface.

53. The level of a liquid, in a capillary tube dipped in a container of the liquid, is different from the general level outside the capillary tube.

54. The difference in levels (h), inside the capillary tube and the container outside, is related to the surface tension $(\mathrm{T})$ of the liquid, by the relation

$$ \mathrm{T}=\dfrac{\mathrm{r}\left(\mathrm{h}+\frac{\mathrm{r}}{3}\right) \rho \mathrm{g}}{2 \cos \theta} $$

Here $r$ is the radius of the capillary tube, $\rho$ is the density of the liquid and $\theta$ is the angle of contact between the liquid and the walls of the capillary tube.

55. We generally use a specially designed ’travelling microscope’ for finding the surface tension of a liquid by the capillary rise method.

Experiment-7: Viscosity

56. Viscosity may be viewed as a sort of ‘internal friction’, between adjacent liquid layers having a ‘relative motion’ between them.

57. According to Newton’s law of viscous flow:

$$ \mathrm{F}=-\eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dz}} $$

( $F=$ viscous force; $A=$ area of layers in contact, $\dfrac{d v}{d z}=$ velocity gradient; $\eta=$ coefficient of viscosity).

58. The Si and CGS units of $\eta$ are $\mathrm{Pa}-\mathrm{s}$ and poise, respectively. We have

$$ 1 \mathrm{~Pa}-\mathrm{s}=10 \text { poise } $$

59. According to Stokes’ law

$$ \mathrm{F}=6 \pi \eta \mathrm{rv} $$

$F=$ viscous force on a spherical ball of radius $r$, when falling through a viscous medium with a velocity v.

60. The ’terminal velocity’, of a falling ball, is that value of its velocity at which the viscous force, on it, just equals its net downward ‘weight’.

61. The formula, for terminal velocity, is

$$ \begin{gathered} 6 \pi \eta \mathrm{rv} _{\mathrm{T}}=\dfrac{4 \pi}{3} \mathrm{r}^{3}(\rho-\sigma) \mathrm{g} \\ \text { or } \quad \mathrm{v} _{\mathrm{T}}=\left[\dfrac{2}{9} \frac{(\rho-\sigma) \mathrm{g}}{\eta}\right] \mathrm{r}^{2} \end{gathered} $$

62. The slope ( $\mathrm{m}$ ), of the graph between $\mathrm{v} _{\mathrm{T}}$ (on $\mathrm{y}$-axis) and $\mathrm{r}^{2}$ (on $x$-axis), can be used to calculate $\eta$ :

$$ \eta=\frac{2(\rho-\sigma) g}{9 m} $$

Experiment-8: Temperature - Time Graph

63. The basic heat transfer formula: Heat gained = Heat lost

64. The transfer of heat takes place through the processes of conduction, convection and radiation.

65. Conduction is important in case of solids; convection is generally the main method of heat transfer in fluids; ‘radiation’ can cause heat transfer even in vacuum.

66. A hot body, kept in ‘isolation’, losses heat mainly through the processes of convection and radiation.

67. The heat energy, emitted through radiation, is quantified through Stefan’s law.

68. According to Stefan’s law:

“The heat radiation, emitted per unit area per second by a perfectly black body, is directly proportional to the fourth power of its absolute (kelvin) temperature”.

$$ \mathrm{E}=\sigma \mathrm{T}^{4} \quad(\sigma=\text { Stefan’s constant }) $$

69. When the (kelvin) temperature of a perfectly black body is close to the (kelvin) temperature of its surroundings, the net heat radiated by it per unit area per second, is (approximately) directly proportional to the excess of its temperature over the surroundings.

70. According to Newton’s law of cooling:

“The rate of fall in temperature, of a hot body, is directly proportional to the difference in temperature between the hot body and its surroundings”.

Thus, $-\dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta _{0}\right)$

71. Newton’s law of cooling leads to the result:

$$ \log \left(\theta-\theta _{0}\right)=-k t+C $$

Thus the graph between $\log \left(\theta-\theta _{0}\right)$ and time $(t)$, has the form shown.

72. It is not quite correct to regard Newton’s law of cooling as an approximation of Stefan’s law.

73. Newtono’s law of cooling is relevant to cooling by convection. When cooling takes place primarily through convection, Newton’s law is (quite) valid even for relatively large difference of temperature between the hot body and its surroundings.

Experiment-9: Resonance Tube

74. Stationary waves are formed through the superposition of a direct wave with its own reflected wave, propagating in the opposite direction.

75. We can get stationary form waves formed in an air column, contained in an open or a closed tube.

76. The permissible modes of vibration of the air column, in an open tube, can have frequencies that are all integral multiples of its fundamental (or simplest) frequency.

77. The permissible modes of vibration, of the air column, in a closed tube, can only have frequencies that are odd integral multiples of its fundamental or simplest frequency.

78. A tuning fork produces a note of a single (constant) frequency that is characterstic of the given tuning fork.

79. The phenomenon of resonance takes place when the frequency of the applied periodic force (used to vibrate a given system) matches the natural frequency of the system itself.

80. The phenomenon of resonance is used in a ‘resonance column set up’ to find the speed of sound in air.

81. The ’end-correction’, associated with the (resonance column) set up, can be eliminated by observing the first and second positions of resonance.

82. The formula used is:

$$ \mathrm{V}=2 \mathrm{n}\left(\ell _{2}-\ell _{1}\right) $$

Experiment-10: Specific Heat Capacity

83. The ‘quantity of heat’ is calculated by using the formula

$$ \mathrm{H}=\mathrm{ms} \theta $$

84. We define the ‘specific heat capacity’, of a substance, as the quantity of heat needed to bring a unit change in the temperature of a unit mass of the substance.

85. The thermal capacity of a substance is the product of its mass and the specific heat capacity of its material.

86. We use the ‘calorimeter’ for doing simple heat related experiments in the laboratory.

87. In the ‘method of mixtures’, the hot body is made to transfer its heat to the cold body that is usually contained in a calorimeter.

88. We can use the formula

$$ \mathrm{s} _{2}=\dfrac{\left[\left(\mathrm{ms}+\mathrm{M} _{1} \mathrm{~s} _{1}\right)\left(\theta _{2}-\theta _{0}\right)\right]}{\left[\mathrm{M} _{2}\left(\theta _{1}-\theta _{2}\right)\right]} $$

for finding the specific heat capacity of a solid. Here ( $\mathrm{m}, \mathrm{M} _{1}, \mathrm{M} _{2}$ ), denote respectively, the mass of calorimeter, the water in it and the solid used.

$\left(\mathrm{s}, \mathrm{s} _{1}, \mathrm{~s} _{2}\right)$, denote, respectively, the specific heat capacities of the material of the calorimeter, water and the solid used.

$\theta _{2}=$ Temperature of hot solid; $\theta _{1}=$ initial temperature of water and calorimeter; $\theta _{2}=$ final equilibrium temperature.

89. We have a similar formula for finding the (unknown) specific heat capacity of a liquid.

Experiment-11: Meter Bridge

90. Wheatstone’s bridge is an arrangment consisting of five resistances $-\mathrm{P}, \mathrm{Q}, \mathrm{R}$ S and $\mathrm{G}-$ which are connected along with a source of electricity, in such a way that no two resistances are in series or in parallel.

91. When the Wheatstone’s bridge is balanced, by adjusting the known variable resistances $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$, the current through the galvanometer (of resistance G), is zero (i.e. $\mathrm{i} _{\mathrm{g}}=0$ ). That gives the condition $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}$, called Wheatstone’s condition for balancing (or Wheatstone’s principle).

92. Wheatstone’s bridge is used to find the value of an unknown resistance (s), usually connected in the fourth arm, from the Wheatstone’s condition for balacing:

$$ \dfrac{\mathrm{P}}{\mathrm{Q}}=\dfrac{\mathrm{R}}{\mathrm{S}} \quad \text { or } \quad \mathrm{S}=\left(\dfrac{\mathrm{Q}}{\mathrm{P}}\right) \mathrm{R} $$

93. Metre bridge is a modified form of Wheatstone’s bridge.

94. Metre bridge is convenient for doing experiments on finding an unknown resistance.

95. The name ‘Metre bridge’ is derived from the fact that it consists of a one ‘metre’ long uniform resistance wire, fixed on a wooden board.

96. Metre bridge is also called slide-wire bridge.

97. The value of the unknown resistance $(\mathrm{X})$ is obtained in terms of the balacing length ’ $\ell$ ‘, as:

$$ X=\dfrac{\ell}{(100-\ell)} \mathrm{R} $$

where $\mathrm{R}$ is the resistance ’taken out’ from the resistance box of the metre bridge circuit.

98. Specific resistance, or resistivity ( $\rho$ ), is a characteristic property of a material; it is a constant, provided the temperature of the material remains consatnt.

99. Specific resistance, of the material of a given wire, of length $L$ and cross-sectional area $A\left(=\dfrac{\pi \mathrm{d}^{2}}{4}\right)$, can be found out, from the resistance $\mathrm{X}$ of this wire through the formula:

$$ \rho=\dfrac{\mathrm{X}}{\mathrm{L}}\left(\frac{\pi \mathrm{d}^{2}}{4}\right) $$

100. For metre bridge, the value of $\mathrm{X}$ is most accurate when the balancing length is, $\ell \simeq 50 \mathrm{~cm}$.

101. The thick metal strips, used in the metre bridge apparatus, provide negligible resistance (almost zero) between verious terminals.

102. Loose contacts, at the terminals, or the plug-keys, will cause error in the determination of X.

Experiment-12: Ohm’s Law

103. Ohm’s law states that current flowing through a conductor is proportional to the potential difference across its ends, when the physical conditions are kept constant.

104. $\mathrm{I}$ $\propto$ $\mathrm{V}$ as per Ohm’s law.

105. $\mathrm{V} \propto \mathrm{I}$ or $\mathrm{V}=\mathrm{RI}$, where $\mathrm{R}$ is resistance of the conductor.

106. Resistance can be viewed as the obstruction offered by the conductor against flow of charges.

107. Resistance is caused due to collisions between moving charges and atoms / molecules of the conductor.

108. Resistance depends on

(a) the nature of the material.

(b) the length of the conductor/wire.

(c) the area of cross-section of the wire.

(d) the temperature of the wire.

109. $\mathrm{R}=\dfrac{\rho \ell}{\mathrm{A}}$

( $\rho=$ specific resistance or resistivity, $\ell=$ length, $A\left(=\dfrac{\pi \mathrm{d}^{2}}{4}\right)$ is area of cross-section).

110. The relation: $\rho _{(t)}=\rho _{(0)}(1+\propto t)$, gives variation of $\rho$ with temeprature $(t) . \propto$ is called the temperature coefficient of resistance / resistivity.

111. $\mathrm{R} _{(\mathrm{t})}=\mathrm{R} _{(0)}(1+\propto \mathrm{t})$ gives the variation of $\mathrm{R}$ with temperature.

112. When the temperature coefficient of resistance is negligible $(\propto \simeq 0)$, $\mathrm{R}$ will not vary (significantly) with temperature.

113. If the resistance of a wire is constant, the graph plotted, for different values of $\mathrm{V}$ across the wire and the corresponding I, will be a straight line.

114. If $\mathrm{V}-\mathrm{I}$ graph is a straight line, the resistor is called an Ohmic resistor.

115. If $\mathrm{V}-\mathrm{I}$ graph is a curve, the resistor is called a non-ohmic resistor.

116. The slope of $\mathrm{V}-\mathrm{I}$ graph (with $\mathrm{V}$ along $\mathrm{y}$-axis and $\mathrm{I}$ along $x$-axis) will give the resistance of the conductor/wire.

Experiment-13: Potentiometer

117. Electrochemical cells, called simply as cells, convert chemical energy into electrical energy.

118. Daniel cell and Leclanche cell are primary cells which can’t be recharged.

119. Lead-acid accumulator, or acid battery, is a secondary cell; it can be recharged.

120. Every cell has got an EMF which is measured not as a force but in terms of work done per unit charge.

121. There are two definitions of EMF of a cell:

(i) In terms of work done: “EMF of a cell is defined as the work done in driving one coulomb of charge through a (closed) circuit, including the cell $\left(\mathrm{E}=\dfrac{\mathrm{W}}{\mathrm{q}}\right)$”

(ii) In terms of terminal voltage $(\mathrm{V})$. “EMF of a cell is defined as the potential difference across the terminals of the cell when no current is known from it”. (i.e. $\mathrm{E}=\mathrm{V}$, when $\mathrm{I}=0$)

122. Internal resistance ( $r$ ) of a cell is the resistance offered by the electrolyte of the cell against the flow of ions through the electrolyte.

123. Internal resistance ( $r$ ) depends on: the temperature and concentration of the electrolyte, the distance between the electrodes and the surface area of the electrodes dipped in the electrolyte.

124. Terminal voltage $(\mathrm{V})$ of a cell is the potential difference across its terminals when current is drawn from it, by an external circuit connected to the cell.

(Terminal voltage is the potential difference across the external circuit).

125. The terminal voltage $(\mathrm{V})$, internal resistance ( $\mathrm{r})$, the $\mathrm{EMF}(\mathrm{E})$ and the external resistance $\mathrm{R}$, across the cell, are related as:

(i) $\mathrm{V}=\dfrac{\mathrm{ER}}{\mathrm{R}+\mathrm{r}}$

or (ii) $r=\left(\dfrac{E}{V}-1\right) R$

126. The principle of potentiometer is that when a steady current (I) is maintained in the uniform wire of the potentiometer, the potential difference $(\mathrm{V})$, obtained across any length $(\ell)$ of the wire, is proportional to the length itself.

i.e. $\mathrm{V} \propto \ell$

or $\mathrm{V}=\mathrm{k} \ell \quad($ where $\mathrm{k}=\mathrm{I} \sigma)$

127. $\mathrm{K}=\dfrac{\mathrm{V}}{\ell}$, is called the potential gradient along the potentiometer wire.

128. Lower is the value of potential gradient, greater will be the accuracy of measurements.

129. The formula used for comparison of EMFs, using potentiometer, is $\dfrac{E _{1}}{E _{2}}=\dfrac{\ell _{1}}{\ell _{2}}$

130. The formula used for determining the internal resistance of a primary cell is:

$$ \mathrm{r}=\left(\dfrac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}, \text { where } \dfrac{\mathrm{E}}{\mathrm{V}}=\dfrac{\ell _{1}}{\ell _{2}} $$

Experiment-14: Moving Coil Galvanometer

131. Galvanometer is an extremely sensitive device, used to detect even very small currents.

132. A galvanometer works on the principle. “A current carrying coil, when suspended in a magnetic field, experiences a torque”.

133. Normally a galvanometer is provided with a linear scale with zero at the centre and with 30 equal divisions on either side of the zero. A pointer, attached to the coil, can deflect and move over the scale, whenever the current passed rotates the coil of the galvanometer.

134. The resistance (G) of the galvanometer is the resistance of the (copper) coil in it. It may (usually) be between $50 \Omega$ and $250 \Omega$, depending on the make and use of the galvanometer.

135. Figure of merit, (k), is the current required to produce one division deflection in the galvanometer. It is given by $\mathrm{k}=\dfrac{\mathrm{I}}{\theta}$

where $\theta$ is the number of divisions of deflection in the galvanometer for a current of I.

136. The unit of k is ampere-divisions ${ }^{-1}$.

137. The formula for finding G, by the ‘half deflection method’, is

$$ \mathrm{G}=\dfrac{\mathrm{RS}}{\mathrm{R}-\mathrm{S}} $$

Here $\mathrm{R}$ is the series resistance which gives a deflection of $\theta$ divisions, when shunt is open. $\mathrm{S}$ is the shunt (parallel) resistance required to reduce the deflection to $\dfrac{\theta}{2}$.

138. The formula for finding $\mathrm{k}$, (the figure of merit, of the galvanometer) is:

$$ \mathrm{k}=\dfrac{\mathrm{E}}{(\mathrm{R}+\mathrm{G}) \theta} $$

Here $E$ is the emf of the cell / battery used, $G$ is value of resistance of the galvanometer (obtained from above formula), $\mathrm{R}$ is series resistance for $\theta$ division deflection.

Experiment-15: Focal Length Measurements

139. Parallex refers to the relative change, in position of one object with respect to another, when both are viewed from difference positions by an observer.

140. Parallex method is used in locating the position of real images, by removing parallex between the real image and a reference pin called the image pin.

(When there is no parallax, the two are at the same positions).

141. A convex mirror also can be made to form a real image by using a convex lens.

142. When the path of a ray is retraced from a convex mirror, the ray will be incident normal on the surface of the mirror. In other words, the ray is directed towards the centre of curvature (C) of the mirror.

143. The formula, used to find focal length ( $f$ ), is $f=\dfrac{R}{2}$, when $R$ is obtained by using the parallax method.

144. Concave mirror can produce real images bigger than the object when object is placed where $u<2 \mathrm{f}$ and $\mathrm{u}>\mathrm{f}$ ). It can produce a smaller real image when $\mathrm{u}>2 \mathrm{f}$.

145. Focal length (f) of concave mirror can be calculated from the formula:

$$ \dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}+\dfrac{1}{\mathrm{u}} $$

146. According to Cartesian sign convention, both $u$ and $v$ are negative (and therefore fis also negative, for concave mirror) when real images are obtained.

147. (u) vs (v) graph, for a concave mirror, is drawn in the third quadrant as both $u$ and $v$ are negative.

148. The scale, for both $u$ and $v$ axes, should be the same and the origin should be at zero (for drawing the (u) vs (v) graph as well as the graph between $\left(\dfrac{1}{u}\right)$ and $\left(\dfrac{1}{v}\right)$.

149. For convex lens, fis positive; it is calculated by using the formula:

$$ \frac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}-\dfrac{1}{\mathrm{u}} $$

150. The (u) vs (v) graph (drawn in the fourth quadrant) and the $\left(\dfrac{1}{u}\right) v s\left(\dfrac{1}{v}\right)$ graph. (drawn in the second quadrant) can also be used for finding $\mathrm{f}$.

151. The scale, for both the axes, should be the same and the origin should be at zero for drawing both (u) vs (v) graph and the $\left(\dfrac{1}{u}\right) \operatorname{vs}\left(\dfrac{1}{v}\right)$ graph for a convex lens also.

Experiment-16: Prism

152. A prism may be regarded as section of a transparent medium, bounded by two plane refracting surfaces meeting at an angle.

This angle $(<\mathrm{A})$ a known as the angle of the prism.

153. When a ray of monochromatic light gets refracted by a prism, we have Angle of incidence + Angle of emergence $=$ Angle of prism + Angle of deviation We also have $<\mathrm{A}=<\mathrm{r} _{1}+<\mathrm{r} _{2}$

154. The graph, between the angle of deviation and the angle of incidence, is a curve which has a point of minima.

155. The minimum value, of the angle of deviation, is known as the angle of minimum deviation.

156. There can be two values of angle of incidence that can correspond to the same value of angle of deviation. (This value of angle of deviation has to be different from its minimum value).

157. One can measure the angle of prism $(<\mathrm{A})$ and find the value of the angle of minimum deviation $\left(<\mathrm{D} _{\mathrm{m}}\right)$. We can then use the relation

$$ \mu=\frac{\sin \left(\dfrac{A+D _{m}}{2}\right)}{\sin \dfrac{A}{2}} $$

to find the refractive index of the material of the prism.

Experiment-17: Glass Slab

158. The phenomena, under the heading ‘refraction’, are due to a change is the speed of light as it goes from one medium to another.

159. The refractive index, ${ } _{1}^{2} \mu$, of medium 2 with respect to medium 1 , equals $\dfrac{\mathrm{v} _{1}}{\mathrm{v} _{2}} \cdot\left(\mathrm{v} _{1}\right.$ and $\mathrm{v} _{2}$ are the speeds of light, in medium 1 and medium 2 , respectively).

160. We also have,as per Snell’s law,

$$ { } _{1}^{2} \mu=\dfrac{\sin \mathrm{i}}{\sin \mathrm{r}} $$

161. The (above) value of refractive index, is also (nearly) equal to the ratio of the ‘real depth’ to the ‘apparent depth’.

162. We can estimate the refractive index, of the material of a given glass slab, by finding its ‘real depth’ and its ‘apparent depth’.

163. The real and apparent depths can be measured by using a ’travelling microscope’.

Experiment-18: p-n Junction Diode

164. A $\mathrm{p}-\mathrm{n}$ junction diode gets ‘forward biased’ when its $\mathrm{p}$-side is at a higher potential than its $\mathrm{n}$-side.

165. It gets ‘reverse biased’ when its $\mathrm{n}$-side is at a higher potential than its $\mathrm{p}$-side.

166. In the forward bias configuration, the current through the diode increases rapidly when the applied voltage goes beyond a certain voltage, known as the ‘knee voltage’ for the given diode.

167. The current, through a p-n junction diode, in its ‘forward bias configuration’, is generally of the order of a few milliamperes; the corresponding current, during its ‘reverse bias configuration’, is generally in the range of a few microamperes.

Experiment-19: ‘Zener Diode’

168. The Zener diode, a special type of $\mathrm{p}-\mathrm{n}$ junction diode, is formed when both the $\mathrm{p}$ and $\mathrm{n}$ sides have been formed through ‘heavy doping’.

169. The breakdown voltage, for a Zener diode, is the voltage, in the reverse bias configuration, at which the (reverse) current suddenly starts increasing quite rapidly.

170. The ‘depletion layer in a Zener diode, is (relatively) very thin. This result in a large electric field, across the junction, when the reverse bias voltage is close to its ‘break down’ voltage value.

171. This large electric field can break the covalent bonds and lead to a phenomenon known as ‘Avalanche breakdown’.

172. The Zener diode is generally used as a ‘voltage regulator’.

Experiment-20: Characterstic Curves of a Transistor

173. Ajunction transistor in a three terminal device; its three terminals are known as the emitter, base and collector.

174. The emitter is forward biased and the collector is reverse biased.

175. We can have three different configurations of junction transistor circuits. The CE (common emitter) configuration is the one used quite often for designing amplifier circuits.

176. The input characterstics, of a transistor in its CE configuration, show the dependence of the base current $\left(\mathrm{I} _{\mathrm{B}}\right)$, on the base emitter voltage $\left(\mathrm{V} _{\mathrm{BE}}\right)$, when the collector emitter voltage is kept constant.

177. The output characterstics, of a transistor in its CE configuration, show the dependence of the collector current $\left(\mathrm{I} _{\mathrm{C}}\right)$ on the collector emitter voltage $\left(\mathrm{V} _{\mathrm{CE}}\right)$, when the base current $\left(\mathrm{I} _{\mathrm{B}}\right)$ is kept constant.

178. We denote the current gain, in the $\mathrm{CE}$ configuration, $\left(=\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{B}}}\right)$ by the symbol $\beta$.

179. For the common base configuration, the current gain $\left(=\dfrac{\Delta \mathrm{I} _{C}}{\Delta \mathrm{I} _{\mathrm{E}}}\right)$, is denoted by the symbol $\alpha$. This ‘current gain’ is always less than 1.

180. We have $\beta=\dfrac{\alpha}{(1-\alpha)}$

Experiment-21: Identifiction of Different Devices

181. We often use a number of devices, (like the diode, LED, transistor, IC’s, resistors and capacitors), in different electronics / electric circuits.

183. Each of these devices has its characterstics features / shape / design.

184. We can use the differences in the features / shape / design, of different devices, to identify them.

Experiment-22: Using the Multimeter

185. The ‘multimeter’ is a versatile measuring / identification instrument; it can be put to a variety of uses.

186. We can use a multimeter for measurement of resistance values as wall as for the measurement of ac/dc currents and voltages.

187. The multimeter can be used to identify the ‘base’ terminal of a transistor. It can also be used to distinguish between npn and pnp transistors.

188. The multimeter also helps us to check the state of a given diode/LED. It does so by making use of the ‘unidirectional flow of current’ through these devices.

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Vernier Callipers

1. When the jaws are just touching each other, the positions of the zeroes of the vernier scale and the main scale, are as shown, for two verniers $A$ and $B$. Vernier $A$ has 20 divisions on its vernier scale and its main scale has been graduated is millimeters.

Vernier B has 50 divisions on its vernier scale and its main scale has been graduated in half millimeters.

Four student reported the following basic information about these two verniers:

	Vcrnicr A		Vcrnicr B
Student	Lcast Count	Zcro Error	Lcast Count	Zcro Error
P	$0.5 \mathrm{~mm}$	$-2.5 \mathrm{~mm}$	$0.1 \mathrm{~mm}$	$+0.5 \mathrm{~mm}$
Q	$0.5 \mathrm{~mm}$	$+2.5 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$-0.05 \mathrm{~mm}$
R	$0.05 \mathrm{~mm}$	$-0.25 \mathrm{~mm}$	$0.1 \mathrm{~mm}$	$+0.5 \mathrm{~mm}$
S	$0.05 \mathrm{~mm}$	$+0.50 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$-0.30 \mathrm{~mm}$

All the results have been correctly quoted by the student

(1) $\mathrm{P}$

(2) $\mathrm{Q}$

(3) $\mathrm{R}$

(4) $\mathrm{S}$

Show Answer

Correct answer: (4)

Solution:

For vernier A:

Least count $=\dfrac{1}{20} \mathrm{~mm}=0.05 \mathrm{~mm}$

Zero error $=+10 \times$ L.C $=0.50 \mathrm{~mm}$

For vernier B:

Least count $=\dfrac{0.5}{50} \mathrm{~mm}=0.01 \mathrm{~mm}$

Zero error $=-30 \times 0.01 \mathrm{~mm}=-0.30 \mathrm{~mm}$

Difficult

Vernier Callipers

2. A given vernier callipers has 20 divisions on its vernier scale which coincide with 19 divisions of the main scale. The main scale has been graduated in $\mathrm{mm}$.

The positions, of the ‘zeros’ of its main scale and its vernier scale, when its jaws just touch one another, are as shown in the figure.

For a given sphere, a student notes that the readings of the main scale, of the above vernier calliper is $2.4 \mathrm{~cm}$ while the $14^{\text {th }}$ division of the vernier coincides with a main scale division. The diameter of this sphere, as per this reading, is

(1) $2.41 \mathrm{~cm}$

(2) $2.53 \mathrm{~cm}$

(3) $2.44 \mathrm{~cm}$

(4) $2.50 \mathrm{~cm}$

Show Answer

Correct answer: (3)

Solution:

The least count of the given vernier $=\dfrac{1}{20} \mathrm{~mm}=0.05 \mathrm{~mm}$

Zero error $=+6 \times$ L.C $=6 \times 0.05 \mathrm{~mm}=0.3 \mathrm{~mm}=0.03 \mathrm{~cm}$

Observed reading,

$$ =\left[2.4+(14 \times 0.05) \times 10^{-1}\right] \mathrm{cm} $$

$=2.47 \mathrm{~cm}$

$\therefore$ Corrected reading $=(2.47-0.03) \mathrm{cm}$

$=2.44 \mathrm{~cm}$

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Vernier Callipers

3. The main scale, on a travelling microscope, has 20 divisions each between its different ‘centimeter marks’. The vernier scale, attached to it, has 50 divisions and these coincide with 49 divisions of the main scale.

The value of one vernier scale division, and the least count of this vernier, are equal, respectively, to

(1) $0.049 \mathrm{~mm}$ and $0.001 \mathrm{~mm}$

(2) $0.049 \mathrm{~mm}$ and $0.001 \mathrm{~cm}$

(3) $0.049 \mathrm{~cm}$ and $0.001 \mathrm{~mm}$

(4) $0.049 \mathrm{~cm}$ and $0.001 \mathrm{~cm}$

Show Answer

Correct answer: (4)

Solution:

There being 20 divisions, between the ‘centimeter marks’, the value of one main scale division is

$$ \left(\dfrac{1}{20}\right) \mathrm{cm}=0.05 \mathrm{~cm} $$

Now 50 division, of the vernier, coincide with 49 division of the main scale. Hence value of one vernier scale division is $\left(\dfrac{49 \times 0.05}{50}\right) \mathrm{cm}=0.049 \mathrm{~cm}$

$\therefore$ Least count $=1$ (M.S.D) -1 (V.S.D) $=(0.05-0.049) \mathrm{cm}$

$$ =0.001 \mathrm{~cm} $$

[Alternatively: Least count $=\dfrac{1 \text { M.S.D }}{\text { No. of divisions on vernier }}$

$$ \left.=\frac{0.05}{50} \mathrm{~cm}=0.001 \mathrm{~cm}\right] $$

Difficult

Vernier Callipers

4. A vernier scale set up, is used to find the internal diameter of a capillary tube. The basic details of this vernier are as follows:

No. of divisions on the vernier scale $=\mathbf{5 0}$

No. of main scale divisions coinciding with these 50 divisions of the vernier scale $=\mathbf{49}$

No. of divisions between the successive centimeter marks on the main scale $=\mathbf{2 0}$

Zero error of this vernier scale $=-(3$ times the L.C $)$

The readings, of the reference line, when the reference line, is in the positions $A$ and $B$, are as follows:

Position	Main scale Reading	Coincident Vernier Division
A	$5.95 \mathrm{~cm}$	7
B	$6.20 \mathrm{~cm}$	3

The (measured) diameter, of this capilary tube, is

(1) $0.255 \mathrm{~cm}$

(2) $0.249 \mathrm{~cm}$

(3) $0.246 \mathrm{~cm}$

(4) $0.252 \mathrm{~cm}$

Show Answer

Correct answer: (3)

Solution:

Value of one M.S.D $=\dfrac{1}{20} \mathrm{~cm}=0.5 \mathrm{~mm}$

No. of divisions on the vernier $=50$

$\therefore \quad$ L.C $=\dfrac{0.5}{50} \mathrm{~mm}=0.01 \mathrm{~mm}=0.001 \mathrm{~cm}$

Zero error $=0.003 \mathrm{~cm}$

Total observed reading at end $\mathrm{A}=(5.95+7 \times 0.001) \mathrm{cm}$

$=5.957 \mathrm{~cm}$

Total observed reading at end $\mathrm{B}$

$=(6.20+3 \times 0.001) \mathrm{cm}$

$=6.203 \mathrm{~cm}$

$\therefore$ Diameter of the capillary tue

$\begin{aligned} & =(6.203-5.957) \mathrm{cm} \\ & =0.246 \mathrm{~cm} \end{aligned}$

[Note: There is no need here to apply ‘zero correction’ to the readings at ends A and B. This is because we have to take the difference of the two readings. The identical correction applied to the two readings, would cancel out in the difference. This can be seen explicity as follows:

Corrected reading at end $\mathrm{A}=[5.957-(-0.003)] \mathrm{cm}$

$=5.960 \mathrm{~cm}$

Corrected reading at end $\mathrm{B}=[6.203-(-0.003)] \mathrm{cm}$

$=6.206 \mathrm{~cm}$

$\therefore$ Diameter $=0.246 \mathrm{~cm}$

The same, as before (when the zero correction was not applied to the two readings).

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Vernier Callipers

5. The positions of the zeroes of the main scale and the vernier scale, when the jaws of the vernier just touch each other, is as shown in the figure.

This vernier is held vertical with the edge of its main scale just touching the mouth of a calorimeter and the end of its sliding strip just touching the bottom of the calorimter.

The reading of the main scale, at this positions is $5.4 \mathrm{~cm}$ and the seventh division, on the vernier scale, coincides with a division on the main scale.

If the main scale is graduated up to $15.0 \mathrm{~cm}$, the depth of the calorimeter is

(1) $5.52 \mathrm{~cm}$

(2) $5.42 \mathrm{~cm}$

(3) $9.48 \mathrm{~cm}$

(4) $9.58 \mathrm{~cm}$

Show Answer

Correct answer: (1)

Solution:

The L.C of the vernier $=\dfrac{1 \mathrm{~mm}}{10}=0.1 \mathrm{~mm}=1.01 \mathrm{~cm}$

Zero error $=-5 \times$ L.C $=-0.05 \mathrm{~cm}$

$=(5.4+7 \times 0.01) \mathrm{cm}=5.47 \mathrm{~cm}$

$\therefore$ Corrected reading $=[5.47-(-0.05)] \mathrm{cm}$

$=5.52 \mathrm{~cm}$

[Note: The (corrected) reading, observed this way, directly gives the depth of the calorimeter].

Average

Screw Guage

6. The positions of the ‘zeroes’ of the circular scale, with respect to the reference lines, (when the jaws are just put in contact), for two screw gauge $X$ and $Y$, are as shown.

Screw gauge $X$ has a main scale graduated in half millimeters and its circular scale has 50 divisions. The screw advances $3.5 \mathrm{~mm}$, on the main scale, when given 7 complete rotations.

‘Screw gauge $Y$ has a main scale graduates in millimeters and its circular scale has 100 divisions. The screw advances $5 \mathrm{~mm}$, on the main scale, when given 5 complete rotations.

Four students reported the following basic information about the two ‘screw gauges’.

Student		Screw Gauge X			Screw Gauge Y
	Pitch	Least Count	Zero Error	Pitch	Least Count	Zero Error
A	$0.5 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$+0.10 \mathrm{~mm}$	$1 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$-0.05 \mathrm{~mm}$
B	$0.5 \mathrm{~mm}$	$0.005 \mathrm{~mm}$	$-0.10 \mathrm{~mm}$	$1 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$+0.05 \mathrm{~mm}$
C	$0.5 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$-0.05 \mathrm{~mm}$	$1 \mathrm{~mm}$	$0.01 \mathrm{mII}$	$+0.05 \mathrm{~mm}$
D	$0.5 \mathrm{~mm}$	$0.005 \mathrm{~mm}$	$+0.05 \mathrm{~mm}$	$1 \mathrm{~mm}$	$0.01 \mathrm{~mm}$	$-0.05 \mathrm{~mm}$

The complete correct reporting has been done by student

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (1)

Solution:

For screw gauge $\mathrm{X}$ :

Pitch $=\dfrac{3.5}{7} \mathrm{~mm}=0.5 \mathrm{~mm}$

Least count $=\dfrac{0.5}{50} \mathrm{~mm}=0.01 \mathrm{~mm}$

Zero error $=+10 \times$ L.C $=+0.10 \mathrm{~mm}$

For screw gauge $\mathrm{Y}$ :

Pitch $=\dfrac{5}{5} \mathrm{~mm}=1 \mathrm{~mm}$

Least count $=\dfrac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}$

Zero error $=-5 \times$ L.C $=-0.05 \mathrm{~mm}$

Average

Screw Guage

7. The main scale, of a given screw gauge, has been graduated in ‘half millimeters’.

The moving jaw of this screw gauge advances through $2.5 \mathrm{~mm}$, on the main scale, when the screw is given five complete rotations. There are 100 divisions on its circular scale.

The zero of the circular scale, coincides with the zero of the main scale when the moving face end just coincides with its fixed face end.

When a thin sheet is just held between the jaws of the screw guage, the main scale reads $0.55 \mathrm{~cm}$ while the $43^{\mathrm{rd}}$ division, of the circular scale, lies along the line of graduations of the main scale. The least count of this screw gauge, and the thickness of the sheet, are equal, respectively, to

(1) $0.0005 \mathrm{~mm}$ and $0.5715 \mathrm{~cm}$

(2) $0.0005 \mathrm{~cm}$ and $0.55215 \mathrm{~cm}$

(3) $0.0005 \mathrm{~mm}$ and $0.55215 \mathrm{~cm}$

(4) $0.0005 \mathrm{~cm}$ and $0.5715 \mathrm{~cm}$

Show Answer

Correct answer: (4)

Solution:

The pitch of the screw gauge $=\dfrac{2.5}{5} \mathrm{~mm}=0.5 \mathrm{~mm}$

$\therefore$ Least count $=\dfrac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}=0.0005 \mathrm{~cm}$

Main scale reading $=0.55 \mathrm{~cm}$

Circular scale reading $=43 \times 0.0005 \mathrm{~cm}$

$=0.0215 \mathrm{~cm}$

$\therefore$ Thickness of the sheet

$=0.5715 \mathrm{~cm}$

Average

Screw Gauge

8. The basic information, about a given screw gauge, is as follows:

(i) Number of divisions on the circular scle $=50$

(ii) Distance moved, by the moving jaw, in 5 complete rotations $=5 \mathrm{~mm}$

(iii) Position of the zero of the circular scale, when the two jaws just touch each other; as shown in the figure

When a wire is just held between the jaws of the screw gauge

(a) Reading of the main scale $=6.2 \mathrm{~mm}$

(b) No. of circular scale divison coinciding with the ‘reference line’ $=38$

The radius, of this wire, equals

(1) $0.326 \mathrm{~cm}$

(2) $0.652 \mathrm{~cm}$

(3) $0.664 \mathrm{~cm}$

(4) $0.332 \mathrm{~cm}$

Show Answer

Correct answer: (1)

Solution:

Pitch $=\dfrac{5}{5} \mathrm{~mm}=1 \mathrm{~mm}=0.1 \mathrm{~cm}$

$\therefore$ Least count $=\dfrac{0.1}{100}=0.001 \mathrm{~cm}$

Zero error $=+6 \times 0.001 \mathrm{~cm}=+0.006 \mathrm{~cm}$

Total observed reading $=(0.62+38 \times 0.001) \mathrm{cm}$

$=0.658 \mathrm{~cm}$

$\therefore$ Corrected reading $=(0.658-0.006) \mathrm{cm}$

$=0.652 \mathrm{~cm}$

Radius of wire $=0.326 \mathrm{~cm}$

Average

Screw Guage

9. The basic information, about a given screw gauge, is as follows:

(i) Least divison on the main scale $=0.5 \mathrm{~mm}$

(ii) No. of divisons on the circular scale $=100$

(iii) Distance moved, on the main scale, when the screw is given 7 complete rotations $=3.5 \mathrm{~mm}$

(iv) Circular scale division, coinciding with the line of graduated (reference) line, when the ‘jaws’, of the screw gauge, are just in contact $=97$

When a wire is just held in contact with the jaws of the screw gauge, the main scale reads $0.45 \mathrm{~mm}$. The circular scale division, just coinciding the reference line, is the $33^{\text {rd }}$ division.

The zero error, and the corrected value of the radius of the wire, are equal, respectively, to

(1) $+0.0015 \mathrm{~cm}$ and $0.0308 \mathrm{~cm}$

(2) $-0.0015 \mathrm{~cm}$ and $0.0315 \mathrm{~cm}$

(3) $-0.0015 \mathrm{~cm}$ and $0.0308 \mathrm{~cm}$

(4) $+0.0015 \mathrm{~cm}$ and $0.0030 \mathrm{~cm}$

Show Answer

Correct answer: (2)

Solution:

Pitch $=\dfrac{3.5}{7} \mathrm{~mm}=0.5 \mathrm{~mm}$

$\therefore \quad$ L.C $=\dfrac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}=0.0005 \mathrm{~cm}$

Zero error $=-(100-97) \times$ L.C $=-(.0015) \mathrm{cm}$

Observed total reading $=(0.045+33 \times 0.0005) \mathrm{cm}$

$=0.0615 \mathrm{~cm}$

Corrected reading $=[0.0615-(-.0015)] \mathrm{cm}=0.0630 \mathrm{~cm}$

$\therefore \quad$ Radius $=0.0315 \mathrm{~cm}$

Average

Screw Gauge

10. The least count, of a screw gauge, having 100 divisons, on its circular scale, equals 0.0005 $\mathrm{cm}$. When a thin sheet in just held between its jaws, the main scale reads $0.65 \mathrm{~mm}$ and the $43^{\text {rd }}$ divison, of the circular scale, coincides with the ‘reference line’. The student, doing the experiment, (correctly), reports the thickness of the sheet as $0.0850 \mathrm{~cm}$.

The pitch and the ‘zero error’ of this screw gauge, are, respectively, equal to

(1) $0.5 \mathrm{~mm}$ and $(+0.0015) \mathrm{cm}$

(2) $0.05 \mathrm{~mm}$ and $(-0.0015) \mathrm{cm}$

(3) $0.5 \mathrm{~mm}$ and $(-0.0015) \mathrm{cm}$

(4) $0.05 \mathrm{~mm}$ and $(+0.0015) \mathrm{cm}$

Show Answer

Correct answer: (1)

Solution:

We have L.C $=\dfrac{\text { Pitch }}{100}$

$\therefore \quad$ Pitch $=0.0005 \times 100 \mathrm{~cm}=0.05 \mathrm{~cm}$

$=0.5 \mathrm{~mm}$

Total observed reading $=(0.65 \mathrm{~mm})+(43 \times 0.0005) \mathrm{cm}$

$=0.0865 \mathrm{~cm}$

Correct reading $=$ Observed reading - (zero error)

$\therefore$ Zero error $=(0.0865-0.0850) \mathrm{cm}$

$=+0.0015 \mathrm{~cm}$

Average

Simple Pendulum

11. The amplitude, of the oscillating bob of a simple pendulum, decreases form an initial value A to a value $(A / n)$ in $T$ seconds. The mass of the bob is $m$ and the time period of the pendulum is $\mathrm{T} _{0}$ seconds.

The average rate of dissipation of energy, by the oscillating bob, is

(1) $\dfrac{2 \pi^{2} \mathrm{~mA}^{2}}{\mathrm{~T}\left(\mathrm{n}^{2} \mathrm{~T} _{0}^{2}\right)}$

(2) $\dfrac{2 \pi^{2} \mathrm{~mA}}{\mathrm{~T}\left(\mathrm{nT} _{0}^{2}\right)}$

(3) $\dfrac{2 \pi^{2} \mathrm{~mA}^{2}}{\mathrm{~T}\left(\mathrm{n}^{2} \mathrm{~T} _{0}^{2}\right)}\left(\mathrm{n}^{2}-1\right)$

(4) $\dfrac{2 \pi^{2} \mathrm{~mA}}{\operatorname{Tn}\left(\mathrm{T} _{0}^{2}\right)}(\mathrm{n}-1)$

Show Answer

Correct answer: (3)

Solution:

Initial Energy $=\dfrac{1}{2} \mathrm{~mA}^{2} \omega^{2}=\dfrac{1}{2} \mathrm{~mA}^{2}\left(\dfrac{2 \pi}{\mathrm{T} _{0}}\right)^{2}$

$=2 \pi^{2} \mathrm{~m} \dfrac{\mathrm{A}^{2}}{\mathrm{~T} _{0}^{2}}$

Final energy $=\dfrac{2 \pi^{2} \mathrm{~m}}{\mathrm{~T} _{0}^{2}}\left(\dfrac{\mathrm{A}^{2}}{\mathrm{n}^{2}}\right)$

$\therefore$ Loss of energy (in time $\mathrm{T})=\dfrac{2 \pi^{2} \mathrm{~m}}{\mathrm{~T} _{0}^{2}} \mathrm{~A}^{2}\left(1-\dfrac{1}{\mathrm{n}^{2}}\right)$

$=\dfrac{2 \pi^{2} \mathrm{~mA}^{2}}{\mathrm{~T} _{0}^{2} \mathrm{n}^{2}}\left(\mathrm{n}^{2}-1\right)$

$\therefore$ Average rate of dissipation of energy

$=\dfrac{2 \pi^{2} \mathrm{~mA}^{2}}{\mathrm{~T}\left(\mathrm{n}^{2} \mathrm{~T} _{0}^{2}\right)}\left(\mathrm{n}^{2}-1\right)$

Difficult

Simple Pendulum

12. A student plots a graph between the ‘square of amplitude’, of the oscillating bob of a simple pendulum and ’time’. This graph is likely to be similar to the graph between the pair of variables, shown in the graph labelled as graph

[Note: For graph A, the values are for an ideal gas.

For graph B the displacement values are for a particle starting from rest and having a uniform acceleration.

For graph $\mathrm{C}$ the height means the (small) height above the surface of earth.

For graph D, $\mathrm{N}$ means the number of radioactive nuclei of a given element. ]

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (4)

Solution:

The ‘square of amplitude’, of the bob, is likely to decay exponentially with time $[A(t)=A(0) \exp (-k t)]$

For graph $A, P \propto \dfrac{1}{\mathrm{~V}}$

For graph $B, \mathrm{s} \alpha \mathrm{t}^{2}$

For graph $C, g(h) \simeq g(0)\left[1-\dfrac{2 h}{R}\right] \quad(R=$ radius of earth)

For graph $D, N(t)=N _{0} \mathrm{e}^{-\lambda t}$

Hence it is only graph in which the variation is an ’exponential decay’ with time.

Average

Simple Pendulum

13. The (small) amplitude, of the oscillating bob, of a simple pendulum becomes $75 \%$ of its initial value in a time of $T$ seconds. The percentage loss in energy, of the bob, in this time, is nearly

(1) $40 \%$

(2) $44 \%$

(3) $52 \%$

(4) $56 \%$

Show Answer

Correct answer: (2)

Solution:

The energy of the bob $\alpha \mathrm{A}^{2}$

$\therefore$ Initial energy $=\mathrm{kA}^{2}$

Final energy $=\mathrm{k}\left(\mathrm{A}\left(\dfrac{75}{100}\right)\right)^{2}$

$=\dfrac{9}{16} \mathrm{kA}^{2}$

$\therefore$ Loss of energy $=\left(1-\dfrac{9}{16}\right) \mathrm{kA}^{2}=\dfrac{7}{16} \mathrm{kA}^{2}$

$\therefore$ Percentage loss of enegy $=\dfrac{\dfrac{7}{16} \mathrm{kA}^{2}}{\left(\mathrm{kA}^{2}\right)} \times 100 \%$

$ =\left(\dfrac{7}{16} \times 100\right) \%=43.75 \%$

$ \approx 44 \%$

Average

Simple Pendulum

14. When a simple pendulum executes small amplitude oscillations in a medium (like air), its amplitude decreases, quite slowly

(1) with time (in a linear way) but its time period remains completely independent of its amplitude.

(2) with time (in a non-linear way) but it time period remains completely independent of its amplitude.

(3) with time (in a non-linear way) and its time period shows a slight increase compared to its time period in free space.

(4) With time (in a linear way) and its time period shows a slight decrease compared to its time period in free space.

Show Answer

Correct answer: (3)

Solution:

The oscillations, of a pendulum bob, in a medium (like air) are damped oscillations. For such damped oscillations, the amplitude decreases exponentially with time, i.e., its decrease is a non-linear one.

The time period, for such damped oscillations, is (slightly) more than its the time period in free space.

Average

Using the Principle of Moments to Find the Mass of a given Meter Scale

15. A meter scale, and a system of three ‘weights’, is set up as shown. The mass of the ‘weight’, labelled as $x$ equals (nearly)

(1) $30 \mathrm{~g}$

(2) $35 \mathrm{~g}$

(3) $40 \mathrm{~g}$

(4) $45 \mathrm{~g}$

Show Answer

Correct answer: (2)

Solution:

Taking moments about the midpoint, $\mathrm{O}$, of the meter scale, we have

$\mathrm{g}[25 \times(50-27)+40(50-38)]$

$=\mathrm{M} _{x} \mathrm{~g}(80-50)$

or $\quad 30 \mathrm{M} _{x}=(575+480)$

$\therefore \mathrm{M} _{x}=\dfrac{1055}{30} \mathrm{~g}=35.17 \mathrm{~g}$

$\approx 35 \mathrm{~g}$

Average

Using the Principle of Moments to Find the Mass of a given Meter Scale

16. A meter scale and two objects, of (known) masses $m _{1}$ and $m _{2}$, are set up as shown. The mass $(m)$ of the meter scale, is

(1) $\left[\dfrac{\mathrm{m} _{1}(\mathrm{a}-\mathrm{b})-\mathrm{m} _{2}(100-\mathrm{a}-\mathrm{c})}{(50-\mathrm{a})}\right]$

(2) $\left[\dfrac{\mathrm{m} _{1}(\mathrm{a}-\mathrm{b})-\mathrm{m} _{2}(50-\mathrm{a}-\mathrm{c})}{(50-\mathrm{a})}\right]$

(3) $\left[\dfrac{\mathrm{m} _{1}(a-b)-\mathrm{m} _{2}(100+\mathrm{a}-\mathrm{c})}{(50+\mathrm{a})}\right]$

(4) $\left[\dfrac{\mathrm{m} _{1}(\mathrm{a}-\mathrm{b})-\mathrm{m} _{2}(50+\mathrm{a}+\mathrm{c})}{(50+\mathrm{a})}\right]$

Show Answer

Correct answer: (1)

Solution:

Total clockwise moments about the wedge

$=\mathrm{m} _{1} \mathrm{~g}(\mathrm{a}-\mathrm{b})$

Total anticlockwise moments about the wedge

$\begin{aligned} & =m g(50-a)+m _{2} g[(50-c)+(50-a)] \\ & =m g(50-a)+m _{2} g(100-a-c) \end{aligned}$

$\therefore$ By principle of moments, we get

$\mathrm{mg}(50-\mathrm{a})+\mathrm{m} _{2} \mathrm{~g}(100-\mathrm{a}-\mathrm{c})=\mathrm{m} _{1} \mathrm{~g}(\mathrm{a}-\mathrm{b})$

or

$\mathrm{m}(50-\mathrm{a})=\mathrm{m} _{1}(\mathrm{a}-\mathrm{b})-\mathrm{m} _{2}(100-\mathrm{a}-\mathrm{c})$

$\therefore \quad \mathrm{m}=\left[\dfrac{\mathrm{m} _{1}(\mathrm{a}-\mathrm{b})-\mathrm{m} _{2}(100-\mathrm{a}-\mathrm{c})}{(50-\mathrm{a})}\right]$

Average

Young’s Modulus

17. A meter scale gets balanced at its $40 \mathrm{~cm}$ mark when two objects, of masses $70 \mathrm{~g}$ and $50 \mathrm{~g}$, are kept, respectively, at its $10 \mathrm{~cm}$ mark and its $80 \mathrm{~cm}$ mark. Without making any other change in this set up, the mass of $50 \mathrm{~g}$ is now replaced by an object of unknown mass. To balance the meter scale now, this object has to be shifted to the $90 \mathrm{~cm}$ mark on the meter scale. The mass of the meter scale, and that of the object used, are, respectively,

(1) $10 \mathrm{~g}$ and $30 \mathrm{~g}$

(2) $15 \mathrm{~g}$ and $45 \mathrm{~g}$

(3) $10 \mathrm{~g}$ and $40 \mathrm{~g}$

(4) $15 \mathrm{~g}$ and $35 \mathrm{~g}$

Show Answer

Correct answer: (3)

Solution:

Let $\mathrm{m} _{0}$ be the mass of the meter scale; we then have, in the first case

$$ \begin{array}{r} \quad 70 \times 30=\mathrm{m} _{0} \times 10+50 \times 40 \\ \therefore \mathrm{m} _{0}=\left(\dfrac{2100-2000}{10}\right)=10 \mathrm{~g} \end{array} $$

In the second case, when the object of unknown mass $((x) g)$, is put at the $90 \mathrm{~cm}$ mark, we have

$$ \begin{aligned} & 70 \times 30=10 \times 10+x \times 50 \\ \therefore \quad & x=\left(\dfrac{2100-100}{50}\right)=40 \mathrm{~g} \end{aligned} $$

Average

Young’s Modulus

18. A force, F, applied to the experimental wire in a Searle’s apparents set up, increases its length by an amount $\Delta \ell$.

If the diameter of this experimental wire were to be doubled and the force applied increased to $2 \mathrm{~F}$, the increase in length would be

(1) $2 \Delta \ell$

(2) $\Delta \ell$

(3) $\dfrac{\Delta \ell}{2}$

(4) $\dfrac{\Delta \ell}{4}$

Show Answer

Correct answer: (4)

Solution:

We have,

$\mathrm{Y}=\dfrac{\dfrac{4 \mathrm{~F}}{\pi \mathrm{d}^{2}}}{\left(\dfrac{\Delta \ell}{\mathrm{L}}\right)}$

Also $\mathrm{Y}=\dfrac{\dfrac{4 \mathrm{~F}}{\pi(2 \mathrm{~d})^{2}}}{\left(\dfrac{\Delta \ell^{\prime}}{\mathrm{L}}\right)}$

$\therefore \dfrac{4 \mathrm{~F}}{\mathrm{~d}^{2}} \times \dfrac{\mathrm{L}}{\Delta \ell}=\dfrac{\mathrm{F}}{\mathrm{d}^{2}} \times \dfrac{\mathrm{L}}{\Delta \ell^{\prime}}$

$\therefore \dfrac{4}{\Delta \ell}=\dfrac{1}{\Delta \ell^{\prime}}$

or $\Delta \ell^{\prime}=\left(\dfrac{\Delta \ell}{4}\right)$

Average

Young’s Modulus

19. A rod of a material, of length $L$ and radius $r$, initially at to room temperature, is held between two rigid supports so that its length does not change with changes in temeperature. The coefficient of linear expansion and the Young’s modulus, of the material of the rod, are $\alpha$ and $\mathbf{Y}$, respectively.

When the temperature of the rod increases by $t^{\circ} \mathrm{C}$, the stress, developed in it, would equal

(1) $\dfrac{Y}{(\alpha t)}$

(2) $(Y \alpha t)$

(3) $(Y \alpha t)\left(\pi r^{2}\right)$

(4) $\dfrac{(Y \alpha t)}{\left(\pi r^{2}\right)}$

Show Answer

Correct answer: (2)

Solution:

When the temperature increases by $\mathrm{t}^{\circ} \mathrm{C}$, the increase in length of the rod is

$\Delta \ell=\mathrm{L} \alpha \mathrm{t}$

The corresponding longitudinal strain is

$\dfrac{\Delta \ell}{L}=\alpha \mathrm{t}$

If Y be the Young’s modulus of the material of the rod, the stress developed is given by

Stress $=\mathrm{Y} \times$ strain $=\mathrm{Y} \alpha \mathrm{t}$

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Poisson’s Ratio

20. A wire, of length $L$ and radius $r$, is subjected to a (longitudinal) force $F$ that causes its length to increase by an amount $\Delta \mathrm{L}$. Assuming that the total volume of the wire remains unchanged even after this increase in its length, the Poisson’s ratio, of the material of the wire, equals

(1) $\left(\dfrac{1}{2}\right)$

(2) $\left(\dfrac{2}{3}\right)$

(3) $\left(\dfrac{3}{4}\right)$

(4) $\left(\dfrac{4}{5}\right)$

Show Answer

Correct answer: (1)

Solution:

The (original) volume of the wire is

$\mathrm{V}=\pi \mathrm{r}^{2} \ell$

When the length of the wire increases by an amount $\Delta \ell$, its radius must decrease (say, by an amount $\Delta \mathrm{r}$ ) so that the volume still equals $\mathrm{V}$. We thus have

$\mathrm{V}=\pi(\mathrm{r}-\Delta \mathrm{r})^{2}(\ell+\Delta \ell)$

We thus have

$\begin{gathered} \quad \mathrm{V}=\pi \mathrm{r}^{2} \ell=\text { constant } \\ \therefore \quad \Delta \mathrm{V}(=0)=\pi(2 \mathrm{r} \Delta \mathrm{r}) \ell+\pi \mathrm{r}^{2}(\Delta \ell) \\ \therefore \quad 2(\Delta \mathrm{r}) \ell+\mathrm{r}(\Delta \ell)=0 \\ \text { or } \quad \dfrac{\Delta \ell}{\ell}=-2\left(\dfrac{\Delta \mathrm{r}}{\mathrm{r}}\right) \end{gathered}$

The Poisson’s ratio, $\sigma$, is given by

$\begin{aligned} \sigma & =\dfrac{-\left(\dfrac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{\left(\dfrac{\Delta \ell}{\ell}\right)}=-\dfrac{\left(\dfrac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{-2\left(\dfrac{\Delta \mathrm{r}}{\mathrm{r}}\right)}=\dfrac{1}{2} \\ \therefore \sigma & =\dfrac{1}{2} \end{aligned}$

Difficult

Young’s Modulus

21. The breaking stress, for a (light) wire of a given material, is $\mathbf{n}$ times the Young’s modulus ( $Y$ ) of its material. The wire has a (normal) length $\ell$ and a radius $r$. A mass $M$ is suspended from its lower free end and is made to oscillate by deflecting the wire through an angle $\theta$ from its normal equilibrium position. The maximum value of $\theta$, for which the stress developed in the wire, at its mean position, would just equal its breaking stress, is

(1) $\cos ^{-1}\left[\dfrac{1}{2 \mathrm{Mg}}\left(2 \mathrm{Mg}-\pi \mathrm{r}^{2} \mathrm{nY}\right)\right]$

(2) $\cos ^{-1}\left[\dfrac{1}{2 \mathrm{Mg}}\left(3 \mathrm{Mg}-2 \pi \mathrm{r}^{2} \mathrm{nY}\right)\right]$

(3) $\cos ^{-1}\left[\dfrac{2}{\mathrm{Mg}}\left(\mathrm{Mg}-\pi \mathrm{r}^{2} \mathrm{nY}\right)\right]$

(4) $\cos ^{-1}\left[\dfrac{1}{2 \mathrm{Mg}}\left(3 \mathrm{Mg}-\pi \mathrm{r}^{2} \mathrm{nY}\right)\right]$

Show Answer

Correct answer: (4)

Solution:

When the wire is displaced through an angle $\theta$, the mass $M$ would rise through a height $h$ where

$\mathrm{h}=(\mathrm{L}-\mathrm{L} \cos \theta)=\mathrm{L}(1-\cos \theta)$

The increase in P.E., of the mass, is Mgh. Hence if $v$ be the velocity, of the mass, at the mean position, we have

$\dfrac{1}{2} \mathrm{Mv}^{2}=\mathrm{Mgh} \quad \text { or } \quad \mathrm{v}^{2}=2 \mathrm{gh}$

At the mean position, the total tension, developed in the wire, would be

$\begin{aligned} \mathrm{T}= & \mathrm{Mg}+\dfrac{\mathrm{Mv}^{2}}{\ell}=\mathrm{Mg}+\dfrac{2 \mathrm{gh} \mathrm{M}}{\ell} \\ & =\mathrm{Mg}+2 \mathrm{Mg} \dfrac{(\mathrm{L}(1-\cos \theta))}{(\mathrm{L})} \\ & =3 \mathrm{Mg}-2 \mathrm{Mg} \cos \theta \end{aligned}$

The corresponding stress is

$\text { Stress }=\dfrac{\mathrm{T}}{\pi \mathrm{r}^{2}}=\dfrac{3 \mathrm{Mg}-2 \mathrm{Mg} \cos \theta}{\pi \mathrm{r}^{2}}$

This would just equal the breaking stress if

$\dfrac{3 \mathrm{Mg}-2 \mathrm{Mg} \cos \theta}{\pi \mathrm{r}^{2}}=\mathrm{nY}$

or $3 \mathrm{Mg}-\pi \mathrm{r}^{2} \mathrm{nY}=2 \mathrm{Mg} \cos \theta$

or $\cos \theta=\dfrac{3}{2}-\dfrac{\pi \mathrm{r}^{2} \mathrm{nY}}{2 \mathrm{Mg}}$

$\therefore \theta=\cos ^{-1}\left[\dfrac{1}{2 \mathrm{Mg}}\left(3 \mathrm{Mg}-\pi \mathrm{r}^{2} \mathrm{nY}\right)\right]$

Average

Surface Tension

22. It is well known that water forms a concave meniscus while mercury forms a convex meniscus when they are present in a glass tube. This is because the

(1) Density of mercury is more than that of glass while water has a density less than that of glass.

(2) Force of cohesion, between water molecules is more than the force of adhesion between water and glass; the reverse in true for mercury.

(3) Force of cohesion, between water molecules, is less thatn the force of adhesion between water and glass; the reverse is true for mercury.

(4) Excess of pressure, inside a water drop of radius $r$, is more than the correspending excess of pressure inside a mercury drop of the same radius.

Show Answer

Correct answer: (3)

Solution:

The shape of the meniscus, of a given liquid, in a given container, is determined primarly by the comparative magnitudes of the relevant forces of cohesion and adhesion.

For liquids (like water), the force of cohesion between liquid molecules, is less than the force of adhesion between the liquid molecules and the molecules of the walls of the container (glass). In such a case, the liquid forms a concave meniscus.

The reverse is true for mercury put in a glass tube. In such a case the liquid forms a convex meniscus.

[Note: Option (4) is incorrect; the excess pressure $\left(=\dfrac{2 \mathrm{~T}}{\mathrm{r}}\right)$ is more for the mercury drop which has a higher value of surface tension].

Average

Capillary Rise

23. Two liquids, (having surface tensions $T _{1}$ and $T _{2}$ and densities $\rho _{1}$ and $\rho _{2}$ ) are made to rise in (small radii) capillary tubes (of glass) of radii $r _{1}$ and $r _{2}$, respectively. Their, respective, angles of contact, with glass, are $\theta _{1}$ and $\theta _{2}$. The first liquid rises to a level $h$ above the general outside level while the second liquid falls to a level $h$ below the general outside level. We would then have

(1) $\dfrac{\cos \theta _{1}}{\cos \theta _{2}}=\dfrac{\left(\mathrm{T} _{1} \mathrm{r} _{1} \rho _{1}\right)}{\left(\mathrm{T} _{2} \mathrm{r} _{2} \rho _{2}\right)}$

(2) $\dfrac{\cos \theta _{1}}{\cos \theta _{2}}=\dfrac{-\left(\mathrm{T} _{2} \mathrm{r} _{1} \rho _{1}\right)}{\left(\mathrm{T} _{1} \mathrm{r} _{2} \rho _{2}\right)}$

(3) $\dfrac{\cos \theta _{1}}{\cos \theta _{2}}=\dfrac{\left(T _{2} r _{2} \rho _{2}\right)}{\left(T _{1} r _{1} \rho _{1}\right)}$

(4) $\dfrac{\cos \theta _{1}}{\cos \theta _{2}}=\dfrac{\left(T _{1} r _{2} \rho _{2}\right)}{\left(T _{2} r _{1} \rho _{1}\right)}$

Show Answer

Correct answer: (2)

Solution:

We know that

$\begin{aligned} & \mathrm{T} \simeq \dfrac{\mathrm{rhg} \rho}{(2 \cos \theta)} \\ & \therefore \mathrm{h}=\dfrac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \end{aligned}$

For the two given liquids

$+\mathrm{h}=\dfrac{2 \mathrm{~T} _{1} \cos \theta _{1}}{\mathrm{r} _{1} \rho _{1} \mathrm{~g}}$

and $-\mathrm{h}=\dfrac{2 \mathrm{~T} _{2} \cos \theta _{2}}{\mathrm{r} _{2} \rho _{2} \mathrm{~g}}$

$\therefore \dfrac{2 \mathrm{~T} _{1} \cos \theta _{1}}{\mathrm{r} _{1} \rho _{1} \mathrm{~g}}=\dfrac{2 \mathrm{~T} _{2} \cos \theta _{2}}{\mathrm{r} _{2} \rho _{2} \mathrm{~g}}$

or $\dfrac{\cos \theta _{1}}{\cos \theta _{2}}=-\dfrac{\mathrm{T} _{2} \mathrm{r} _{1} \rho _{1}}{\mathrm{~T} _{1} \mathrm{r} _{2} \rho _{2}}$

Average

Capillary Rise

24. A given liquid rises to a height $h$ in a given capillary tube. If this capillary tube is cut to have a length $\ell$ that is less than $\mathrm{h}$, the liquid

(1) will start over flowing through the capillary tube.

(2) will have a readjusted value of its surface tension for which the rise is equal to $\ell$.

(3) will readjust its angle of contact, with the walls of the container, to a new value, say $\theta^{\prime}$, where

$\theta^{\prime}=\cos ^{-1}\left[\dfrac{\mathrm{r} \ell \rho \mathrm{g}}{2 \mathrm{~T}}\right]$

(4) will readjust its angle of contact, with the walls of the container, to a new value, say $\theta$ “, where

$\theta^{\prime \prime}=\cos ^{-1}\left[\dfrac{2 \mathrm{~T}}{\mathrm{r} \ell \rho \mathrm{g}}\right]$

Show Answer

Correct answer: (3)

Solution:

The liquid will not overflow as that would lead to a violation of the principle of conservation of energy. For the rise (h) of a liquid, in a capillary tube of radius, $\mathrm{r}$, we have the relation

$$ \mathrm{T}=\dfrac{\mathrm{r}\left(\mathrm{h}+\dfrac{\mathrm{r}}{3}\right) \rho \mathrm{g}}{2 \cos \theta} \simeq \dfrac{\mathrm{rh} \rho \mathrm{g}}{2 \cos \theta} $$

For a capillary tube, of insufficient height, the liquid, readjusts its angle of contact with the walls of the container, so that

$$ \cos \theta^{\prime}=\dfrac{(\mathrm{r} \ell) \rho \mathrm{g}}{2 \mathrm{~T}} \text { or } \quad \theta^{\prime}=\cos ^{-1}\left[\dfrac{(\mathrm{r} \ell) \rho \mathrm{g}}{2 \mathrm{~T}}\right] $$

[Note: The liquid does not readjust its surface tension to a new value].

Difficult

Capillary Rise

25. Two liquids (say $L _{1}$ and $L _{2}$ ), having surface tensions $T _{1}$ and $T _{2}$ (where $\left(\dfrac{T _{1}}{T _{2}}\right)=\dfrac{2}{3}$ ) and densities $\rho _{1}$ and $\rho _{2}$ (where $\left(\dfrac{\rho _{1}}{\rho _{2}}\right)=\dfrac{1}{2}$ ), rise to the same height (say h) (above their respective outside levels) in identical (very small radius) capillary tubes. If their respective angles of contact, with the walls of the capillary tubes, are $\theta _{1}$ and $\theta _{2}$, we would have

(1) $9 \cos ^{2} \theta _{2}+16 \cos ^{2} \theta _{2}+7=0$

(2) $16 \cos ^{2} \theta _{2}-9 \cos ^{2} \theta _{1}-7=0$

(3) $9 \sin ^{2} \theta _{1}+16 \sin ^{2} \theta _{2}+7=0$

(4) $16 \sin ^{2} \theta _{1}-9 \sin ^{2} \theta _{2}-7=0$

Show Answer

Correct answer: (4)

Solution:

We know that

$\begin{array}{r} \mathrm{T}=\dfrac{\mathrm{rh} \rho \mathrm{g}}{2 \cos \theta} \\ \therefore \quad \mathrm{h}=\dfrac{(2 \mathrm{~T} \cos \theta)}{(\mathrm{r} \rho \mathrm{g})} \end{array}$

From the given data, we then have

$\begin{aligned} & \mathrm{h}=\dfrac{2 \mathrm{~T} _{1} \cos \theta _{1}}{\mathrm{r} \rho _{1} \mathrm{~g}}=\dfrac{2 \mathrm{~T} _{2} \cos \theta _{2}}{\mathrm{r} \rho _{2} \mathrm{~g}} \\ & \therefore \quad \dfrac{\cos \theta _{1}}{\cos \theta _{2}}=\left(\dfrac{\mathrm{T} _{2}}{\mathrm{~T} _{1}}\right) \times\left(\dfrac{\rho _{1}}{\rho _{2}}\right)=\dfrac{3}{2} \times \dfrac{1}{2}=\dfrac{3}{4} \\ & \therefore \quad \dfrac{\cos ^{2} \theta _{1}}{\cos ^{2} \theta _{2}}=\dfrac{9}{16} \\ & \text { or } 16 \cos ^{2} \theta _{1}-9 \cos ^{2} \theta _{2}=0 \\ & \text { or } 16\left(1-\sin ^{2} \theta _{1}\right)-9\left(1-\sin ^{2} \theta _{2}\right)=0 \\ & \text { or } \quad-16 \sin ^{2} \theta _{1}+9 \sin ^{2} \theta _{2}+7=0 \\ & \text { or } 16 \sin ^{2} \theta _{1}-9 \sin ^{2} \theta _{2}-7=0 \end{aligned}$

$\therefore$ It is option (4) that is the correct choice.

Average

Terminal Velocity

26. A drop of a liquid, of radius $r$, acquires a terminal velocity, $v _{0}$, when falling vertically through a given viscous medium. In $\mathbf{n}$ such (identical) drops of this liquid were to coalesce to form a single drop, the terminal velocity, $\mathbf{V}$, of this (large) drop, when falling vertically through the same viscous medium, would be

(1) $\mathrm{V}=\mathrm{nv} _{0}$

(2) $\mathrm{V}=\mathrm{n}^{2 / 3} \mathrm{v} _{0}$

(3) $\mathrm{V}=\mathrm{n}^{1 / 3} \mathrm{v} _{0}$

(4) $\mathrm{V}=\mathrm{n}^{-1 / 3} \mathrm{v} _{0}$

Show Answer

Correct answer: (2)

Solution:

Let $\mathrm{R}$ be the radius of the large drop.

We then have

$\dfrac{4 \pi}{3} R^{3}=n\left(\dfrac{4 \pi}{3} r^{3}\right)$

$\therefore \mathrm{R}=\left(\mathrm{n}^{1 / 3}\right) \mathrm{r}$

The relation, between the terminal velocity and the radius of the drop, is given by

$6 \pi \eta \mathrm{rv} _{\mathrm{T}}=\dfrac{4 \pi}{3} \mathrm{r}^{3}(\rho-\sigma) \mathrm{g}$

or $\quad \mathrm{v} _{\mathrm{T}}=\dfrac{2}{9}(\rho-\sigma) \mathrm{g} \cdot \mathrm{r}^{2}$

$\therefore \dfrac{\mathrm{V}}{\mathrm{v} _{0}}=\dfrac{\mathrm{R}^{2}}{\mathrm{r}^{2}}=\dfrac{\left(\mathrm{n}^{1 / 3} \mathrm{r}\right)^{2}}{\mathrm{r}^{2}}=\mathrm{n}^{2 / 3}$

$\therefore \mathrm{V}=\mathrm{n}^{2 / 3} \mathrm{r}$

Difficult

Terminal Velocity

27. A drop of a liquid, of a given radius, falls vertically through a given viscous medium. The pair of graphs, that respresent the variation of velocity (v), and acceleration (a) of the falling drop, with time, is the pair marked as pair

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (4)

Solution:

As the drop falls vertically through the viscous medium, its velocity keeps on increasing with time. This, in turn, increases the viscous force acting on it so that its acceleration keeps on decreasing with time. When the viscous force becomes large enought to balance the net downward force, the acceleration of the drop becomes zero. Its velocity than remains constant (with time) at a value called the terminal velocity of the graph.

This variation of velocity and acceleration, with time, is non-linear because of the non-uniform accleration of the falling drop. The pair (D) shows the variations in their correct form.

Average

Terminal Velocity

28. A drop of a liquid, of density $\rho _{1}$, falls through a medium of density $\sigma$ and ceofficient of viscosity. Another drop, of a liquid of density $\rho _{2}$, having the same radius as that of the first drop, falls through the same medium. It is given that

$$ \rho _{2}=n^{2} \rho _{1} ; \sigma=\left(1-\dfrac{1}{n}\right) \rho _{1} \text { and } \rho _{2}»\sigma $$

The terminal velocities ( $v _{1}$ and $v _{2}$ ) of the first, and the second drops, will then be related as

(1) $\mathrm{v} _{2}=\dfrac{\left(\mathrm{n}^{2}-1\right)}{\mathrm{n}} \mathrm{v} _{1}$

(2) $\mathrm{v} _{2}=\left(\dfrac{\mathrm{n}^{2}}{\mathrm{n}-1}\right) \mathrm{v} _{1}$

(3) $\mathrm{v} _{2} \simeq \mathrm{n}^{3} \mathrm{v} _{1}$

(4) $\mathrm{v} _{2}=\mathrm{n}^{2} \mathrm{v} _{1}$

Show Answer

Correct answer: (3)

Solution:

We know that

$\dfrac{4 \pi}{3} r^{3}(\rho-\sigma) g=6 \pi \eta \mathrm{rv} _{\mathrm{T}}$

$\therefore \quad$ For the first drop

$\dfrac{4 \pi}{3} r^{3}\left[\rho _{1}-\left(1-\dfrac{1}{n}\right) \rho _{1}\right] g=6 \pi \eta r _{1}$

and for the second drop,

$\begin{aligned} & \dfrac{4 \pi}{3} \mathrm{r}^{3}\left[\rho _{2}-\sigma\right] \mathrm{g}=6 \pi \eta \mathrm{rv} _{2} \\ & \text { or } \quad \dfrac{4 \pi}{3} \mathrm{r}^{3} \rho _{2} \simeq 6 \pi \eta \mathrm{rv} _{2} \quad\left(\because \rho _{2}»\sigma\right) \\ & \therefore \quad \dfrac{\rho _{2}}{\rho _{1}} \mathrm{n} \simeq \dfrac{\mathrm{v} _{2}}{\mathrm{v} _{1}} \\ & \text { or } \mathrm{n}^{2} \cdot \mathrm{n} \simeq \dfrac{\mathrm{v} _{2}}{\mathrm{v} _{1}} \\ & \therefore \quad \mathrm{v} _{2} \simeq \mathrm{n}^{3} \mathrm{v} _{1} \end{aligned}$

Difficult

Stoke’s Law

29. A spherical ball, of mass $m$ and radius $r$, starts from rest and falls vertically through a medium of negligible density and having a coefficient of viscosity $\eta$. The time (T) taken by the ball to acquire a velocity $V$, is given by

(1) $\mathrm{T}=\dfrac{6 \pi \eta \mathrm{r}}{\mathrm{m}} \ell \mathrm{n}\left[\dfrac{\mathrm{mg}}{\mathrm{mg}-6 \pi \eta \mathrm{rV}}\right]$

(2) $\mathrm{T}=\dfrac{\mathrm{m}}{6 \pi \eta \mathrm{r}} \ln \left[\dfrac{\mathrm{mg}+6 \pi \eta \mathrm{rV}}{\mathrm{mg}}\right]$

(3) $\mathrm{T}=\dfrac{6 \pi \eta \mathrm{r}}{\mathrm{m}} \ell \mathrm{n}\left[\dfrac{\mathrm{mg}+6 \pi \eta \mathrm{rV}}{\mathrm{mg}}\right]$

(4) $\mathrm{T}=\dfrac{\mathrm{m}}{6 \pi \eta \mathrm{r}} \ell \mathrm{n}\left[\dfrac{\mathrm{mg}}{\mathrm{mg}-6 \pi \eta \mathrm{rV}}\right]$

Show Answer

Correct answer: (4)

Solution:

Let the velocity of the ball be $v$ after a time $t$. The net downward force, acting on to ball, at this instant, is

$\mathrm{F}(\mathrm{t})=\mathrm{mg}-6 \pi \eta \mathrm{rv}$

The acceleration, $a(t)$, at this instant, is

$\begin{aligned} & \mathrm{a}(\mathrm{t})=\dfrac{\mathrm{F}(\mathrm{t})}{\mathrm{m}}\left[\left(\mathrm{g}-\dfrac{6 \pi \eta \mathrm{r}}{\mathrm{m}}\right) \mathrm{v}\right]=(\mathrm{g}-\mathrm{kv}) \\ & \left(\mathrm{k}=\dfrac{6 \pi \eta \mathrm{r}}{\mathrm{m}}\right) \end{aligned}$

$\therefore \dfrac{\mathrm{dv}}{\mathrm{dt}}=(\mathrm{g}-\mathrm{kv})$

or $\dfrac{d v}{(g-k v)}=d t$

$\therefore \quad \int _{0}^{\mathrm{T}} \mathrm{dt}=\int _{0}^{\mathrm{V}} \dfrac{\mathrm{dv}}{(\mathrm{g}-\mathrm{kv})}=-\dfrac{1}{\mathrm{k}}[\ln (\mathrm{g}-\mathrm{kv})] _{0}^{\mathrm{V}}$

$\therefore \quad \mathrm{T}=\dfrac{1}{\mathrm{k}} \ln \left(\dfrac{\mathrm{g}}{\mathrm{g}-\mathrm{kV}}\right)$

or $\quad \mathrm{T}=\left(\dfrac{\mathrm{m}}{6 \pi \eta \mathrm{r}}\right)\left[\ln \left(\dfrac{\mathrm{mg}}{\mathrm{mg}-6 \pi \eta \mathrm{rV}}\right)\right]$

Average

Newton’s Law of Cooling

30. To do an experimental verification of Newton’s law of cooling, we need to take the ’temperature’ vs ’time’ readings, of a hot body, kept in given surroundings. An appropriate choice, of the ‘hot body’ and the ‘surroundings’, would then be:

(1) [A hot metal sphere]; and [a ‘closed enclosure’].

(2) [A hot metal sphere]; [a stone plate to hold the sphere and kept in an uncovered enclosure].

(3) [An uncovered calorimeter containing hot water that is constantly being stirred]; [a slab on an open window or a table under a fan].

(4) [A fully covered calorimeter containing hot water]; [a corner of the room away from any open window or a fan].

Show Answer

Correct answer: (3)

Solution:

Newton’s law of coolings holds for the cooling of a hot body, primarily by convection. It can then be valid even for quite a large difference of temperature between the hot body and the surroundings.

The ‘set-up’, suggested in option (3), is most likely to ensure that the whole of the hot body cools primarily by convection. Hence it is the appropriate choice for verifying Newton’s law of cooling.

Difficult

Cooling Time

31. A hot body, initially at a temperature $8 \theta$, is present in surroundings that are at a temperature $\theta$. The time intervals, for its temperature to fall from, (i) $8 \theta$ to $4 \theta$ and (ii) $4 \theta$ to $2 \theta$, are $\Delta t _{1}$ and $\Delta t _{2}$, respectively. The ratio $\left(\dfrac{\Delta t _{1}}{\Delta t _{2}}\right)$ equals

(1) 2

(2) $\log 2$

(3) $\dfrac{\log ^{(7 / 3)}}{\log (3)}$

(4) $\left[\dfrac{\log 7}{\log 2}\right]$

Show Answer

Correct answer: (3)

Solution:

We know that

$-\dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta _{0}\right)$

$\therefore \dfrac{\mathrm{d} \theta}{\left(\theta-\theta _{0}\right)}=-\mathrm{kdt}$

$\therefore \int \limits _{\theta_1}^{\theta_2} \dfrac{d \theta}{\left(\theta-\theta_0\right)}=-k \int \limits _{t_1}^{t_2} d t$

$\therefore\left|\ell \mathrm{n}\left(\theta-\theta _0\right)\right| _{\theta_1}^{\theta_2}=-\mathrm{k}\left(\mathrm{t}_2-\mathrm{t}_1\right)$

or $\quad \ell \mathrm{n}\dfrac{\left(\theta _{1}-\theta _{0}\right)}{\left(\theta _{2}-\theta _{0}\right)}=\mathrm{k}\left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)$

$\therefore \quad$ (i) $\quad \ell \mathrm{n}\dfrac{(8 \theta-\theta)}{(4 \theta-\theta)}=\mathrm{k}(\Delta \mathrm{t}) _{1}$

and (ii) $\ell \mathrm{n}\dfrac{(4 \theta-\theta)}{(2 \theta-\theta)}=\mathrm{k}(\Delta \mathrm{t}) _{2}$

$\therefore \dfrac{(\Delta \mathrm{t})_1}{(\Delta \mathrm{t})_2}=\dfrac{\ell \mathrm{n}\left(\dfrac{7}{3}\right)}{\ell \mathrm{n}(3)}=\dfrac{\log (1 / 3)}{\log (3)}$

Average

Graph of Increase of Temperature with Time

32. A given object (initially at a temperature $\theta _{0}$ ) is being heated by a ‘heat source’ that supplies it heat at a constant rate. The graph, showing the variation of temperature of the object, with time, would then be the graph labelled as graph

(1) $\mathrm{A}$

(2) $\mathrm{B}$

(3) $\mathrm{C}$

(4) $\mathrm{D}$

Show Answer

Correct answer: (2)

Solution:

Let the temperature of the object (mass $=m$; specific heat capcity $=\mathrm{s}$ ), rise by an amount $\Delta \theta$ when it receives an amount of heat $\Delta \mathrm{H}$. We then have

$\begin{aligned} & \Delta H=m s \Delta \theta \\ \therefore \quad & \dfrac{\Delta H}{\Delta t}=\mathrm{ms} \dfrac{\Delta \theta}{\Delta t} \\ \therefore \quad & \dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\dfrac{1}{\mathrm{~ms}} \dfrac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{a} \text { constant }=\mathrm{k}(\text { say }) \\ \therefore \quad & \mathrm{d} \theta=\mathrm{kdt} \end{aligned}$

$\text { or }\int \limits _{\theta_0}^\theta \mathrm{d} \theta=\mathrm{k} \int \limits _0^{\mathrm{k}} \mathrm{dt} \quad\left(\theta_0=\text { initial temperature }\right)$

$\therefore \quad \theta=\theta _{0}+\mathrm{kt}=\mathrm{kt}+\theta _{0}$

$\therefore$ The graph of $\theta$, against $\mathrm{t}$, is a straight line with a slope $\mathrm{k}$ and an intercept (on the $\theta$-axis) of $\theta _{0}$.

Average

Experimental Skills

33. A hot body, initially at a temperature $\theta$, is kept in surroundings that are at a temperature $\theta _{0}$. The graph, shown here, is a graph of a function, $f(\theta)$, of $\theta$, against time. Assuming that the body loses heat primarily by convection, the function $f(\theta)$, is

(1) $\mathrm{f}(\theta)=\mathrm{k}\left(\theta-\theta _{0}\right)$

(2) $f(\theta)=\dfrac{k}{\left(\theta-\theta _{0}\right)}$

(3) $f(\theta)=\dfrac{k}{\log \left(\theta-\theta _{0}\right)}$

(4) $f(\theta)=k \log \left(\theta-\theta _{0}\right)$

Show Answer

Correct answer: (4)

Solution:

As per Newton’s law of cooling

$-\dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}^{\prime}\left(\theta-\theta _{0}\right)$

or $-\dfrac{\mathrm{d} \theta}{\left(\theta-\theta _{0}\right)}=\mathrm{k}^{\prime} \mathrm{dt}$

$\ell \mathrm{n}\left(\theta-\theta _{0}\right)=-\mathrm{k}^{\prime} \mathrm{t}+\mathrm{C}^{\prime}$

or $\log \left(\theta-\theta _{0}\right)=-\mathrm{kt}+\mathrm{C}^{\prime}$

$\therefore$ The graph of $\log \left(\theta-\theta _{0}\right)$, against time, is a straight line having a negative slope. The given graph has this very form.

Hence $\mathrm{f}(\theta)=\log \left(\theta-\theta _{0}\right)$

Easy

Speed of Sound in Air

34. A student sets up the speed of sound in air apparatus for finding the velocity of sound in air, at the room temperature. Using a tuning fork, of frequency $n(\mathrm{~Hz})$, she obtains the first and second positions of resonance for lengths $\ell _{1}(\mathbf{m})$ and $\ell _{2}(\mathbf{m})$, respectively, of the air column, in the resonance tube. The velocity of sound, $V\left(\right.$ in $\left.\dfrac{m}{s}\right)$, would then be best given by the formula.

(1) $\mathrm{V}=4 \ell _{1} \mathrm{n}$

(2) $\mathrm{V}=\dfrac{4}{3} \ell _{2} \mathrm{n}$

(3) $\mathrm{V}=2\left(\ell _{2}-\ell _{1}\right) \mathrm{n}$

(4) $\mathrm{V}=\left(\ell _{2}+\ell _{1}\right) \mathrm{n}$

Show Answer

Correct answer: (3)

Solution:

We have

$\ell _{1}+\mathrm{e}=\dfrac{\lambda}{4}$

and $\quad \ell _{2}+\mathrm{e}=\dfrac{3 \lambda}{4}$

$\therefore \quad \ell _{2}-\ell _{1}=\left(\dfrac{3 \lambda}{4}-\dfrac{\lambda}{4}\right)=\dfrac{\lambda}{2}$

$\therefore \lambda=2\left(\ell _{2}-\ell _{1}\right)$

The velocity, $\mathrm{V}$, is, therefore, given by

$\begin{aligned} & \mathrm{V}=\mathrm{n} \lambda \\ & =2 \mathrm{n}\left(\ell _{2}-\ell _{1}\right) \end{aligned}$

[Note: It is only this formula for $\mathrm{V}$ that is independent of the ’end-correction’].

Average

Speed of Sound in Air

35. The experiment, using a resonance tube set-up, was carried out in a room at a temperature $\theta^{0} \mathrm{C}$. The tuning fork used had a frequency $\mathrm{N}(\mathrm{Hz})$ and the first and second positions of resonance were observed for lengths $\ell _{1}(\mathrm{~cm})$ and $\ell _{2}(\mathrm{~cm})$, respectively, of the air-column.

From these observations, one can say that the velocity of sound, in air, at $0^{\circ} \mathbf{C}$, equals

(1) $\dfrac{\mathrm{N}\left(\ell _{2}-\ell _{1}\right)}{50}\left(1+\dfrac{\theta}{273}\right)^{1 / 2}(\mathrm{~m} / \mathrm{s})$

(2) $\dfrac{\mathrm{N}\left(\ell _{2}-\ell _{1}\right)}{50}\left(1+\dfrac{\theta}{273}\right)^{-1 / 2}(\mathrm{~m} / \mathrm{s})$

(3) $2 \mathrm{~N}\left(\ell _{2}-\ell _{1}\right)\left(1+\dfrac{\theta}{273}\right)^{-1 / 2}(\mathrm{~m} / \mathrm{s})$

(4) $2 \mathrm{~N}\left(\ell _{2}-\ell _{1}\right)\left(1+\dfrac{\theta}{273}\right)^{1 / 2}(\mathrm{~m} / \mathrm{s})$

Show Answer

Correct answer: (2)

Solution:

We have

$\ell _{1}+\mathrm{e}=\dfrac{\lambda}{4}$

and $\quad \ell _{2}+\mathrm{e}=\dfrac{3 \lambda}{4}$

$\therefore \lambda=\dfrac{1}{50}\left(\ell _{2}-\ell _{1}\right)$ metre

$\therefore \mathrm{V}\left(\right.$ at $\left.\theta^{0} \mathrm{C}\right)=\mathrm{N} \lambda=\dfrac{\mathrm{N}\left(\ell _{2}-\ell _{1}\right)}{50} \mathrm{~m} / \mathrm{s}$

We also have

$\begin{aligned} & \dfrac{\mathrm{V} _{\theta}}{\mathrm{V} _{0}}=\left(\dfrac{\theta+273}{273}\right)^{1 / 2}=\left(1+\dfrac{\theta}{273}\right)^{1 / 2} \\ \therefore & \mathrm{V} _{\theta}=\mathrm{V} _{0}\left(1+\dfrac{\theta}{273}\right)^{-1 / 2} \\ \therefore & \mathrm{V} _{0}=\left[\dfrac{\mathrm{N}}{50}\left(\ell _{2}-\ell _{1}\right)\left(1+\dfrac{\theta}{273}\right)^{-1 / 2}\right] \mathrm{ms}^{-1} \end{aligned}$

Average

Speed of Sound in Air

36. Two students, working in the same laboratory, do the experiment on finding the velocity of sound using the ‘resonance tube column apparatus’. They used tubes of diameters $D _{1}$ and $D _{2}$ and identical tuning forks of frequency $\mathrm{N}$ each. The first and second resonance lengths, found by them, were $\left(\ell _{1}, \ell _{1}^{\prime}\right)$ and $\left(\ell _{2}, \ell _{2}^{\prime}\right)$, respectively. These lengths would be related to each otehr as

(1) $\left(\ell _{1}^{\prime}=\ell _{1}\right) ;\left(\ell _{2}^{\prime}=\ell _{2}\right)$

(2) $\left(\ell _{1}^{\prime}=\ell _{1}\right) ; \ell _{2}^{\prime}=\left[\ell _{2}+0.3\left(\mathrm{D} _{2}-\mathrm{D} _{1}\right)\right]$

(3) $\left[\ell _{1}^{\prime}=\ell _{1}+0.3\left(\mathrm{D} _{1}-\mathrm{D} _{2}\right)\right] ;\left[\ell _{2}^{\prime}=\ell _{2}+0.3\left(\mathrm{D} _{1}-\mathrm{D} _{2}\right)\right]$

(4) $\left[\ell _{1}^{\prime}=\ell _{1}+0.3\left(\mathrm{D} _{2}-\mathrm{D} _{1}\right)\right] ;\left[\ell _{2}^{\prime}=\ell _{2}+0.9\left(\mathrm{D} _{2}-\mathrm{D} _{1}\right)\right]$

Show Answer

Correct answer: (3)

Solution:

We know that

$\dfrac{\lambda}{4}=$ first resonance length + end correction

and $\dfrac{3 \lambda}{4}=$ second resonance length + end correction

Also end correction $=0.3$ (Diameter of the tube). As the students are working in the same laboratory (hence same value of velocity of sound) and are using tuning forks of same frequency, the value of $\lambda$ must be the same in both cases $\left(\lambda=\dfrac{\mathrm{V}}{\mathrm{N}}\right)$. We, therefore, have

$\dfrac{\lambda}{4}=\left(\ell _{1}+0.3 \mathrm{D} _{1}\right)=\left(\ell _{1}^{\prime}+0.3 \mathrm{D} _{2}\right)$

$\therefore\left(\ell _{1}^{\prime}-\ell _{1}\right)=0.3\left(\mathrm{D} _{1}-\mathrm{D} _{2}\right)$

and $\dfrac{3 \lambda}{4}=\left(\ell _{2}+0.3 \mathrm{D} _{1}\right)=\left(\ell _{2}^{\prime}+0.3 \mathrm{D} _{2}\right)$

$\therefore \quad\left(\ell _{2}^{\prime}-\ell _{2}\right)=0.3\left(\mathrm{D} _{1}-\mathrm{D} _{2}\right)$

Difficult

Speed of Sound in Air

37. A student did the experiment on finding the speed of sound (say $V$ ) using a given resonance tube of diameter $D$. She used five tuning forks of different frequencies and observed the first and second resonance lengths in each case. She then plotted (separate) graphs of these first and second resonance lengths against the reciprocal of the corresponding tuning fork frequencies. The slopes of the two graphs were found to be $m _{1}$ and $m _{2}$ while the magnitudes of their intercepts, on the $y$-axis, were found to be $C _{1}$ and $C _{2}$, respectively. We would then have

(1) $\mathrm{V}=4 \mathrm{~m} _{1}=\dfrac{4}{3} \mathrm{~m} _{2}$ and $\mathrm{D}=\dfrac{10}{3}\left|\mathrm{C} _{1}\right|=\dfrac{10}{3}\left|\mathrm{C} _{2}\right|$

(2) $\mathrm{V}=4 \mathrm{~m} _{1}=\dfrac{3}{4} \mathrm{~m} _{2}$ and $\mathrm{D}=\dfrac{10}{3}\left|\mathrm{C} _{1}\right|=10\left|\mathrm{C} _{2}\right|$

(3) $\mathrm{V}=4 \mathrm{~m} _{1}=\dfrac{3}{4} \mathrm{~m} _{2} \quad \text { and } \quad \mathrm{D}=10\left|\mathrm{C} _{1}\right|=\dfrac{10}{3}\left|\mathrm{C} _{2}\right|$

(4) $\mathrm{V}=\dfrac{\mathrm{m} _{1}}{4}=\dfrac{3}{3} \mathrm{~m} _{2}$ and $\mathrm{D}=10\left|\mathrm{C} _{1}\right|=\dfrac{10}{3}\left|\mathrm{C} _{2}\right|$

Show Answer

Correct answer: (1)

Solution:

We know that

$\mathrm{V}=\mathrm{N} .4\left(\ell _{1}+\mathrm{e}\right)$

or $\quad \ell _{1}=\dfrac{\mathrm{V}}{4} \cdot \dfrac{1}{\mathrm{~N}}-\mathrm{e}$

Also $\mathrm{V}=\mathrm{N} \cdot \dfrac{4}{3}\left(\ell _{2}+\mathrm{e}\right)$

or $\quad \ell _{2}=\left(\dfrac{3}{4} \mathrm{~V}\right) \cdot \dfrac{1}{\mathrm{~N}}-(\mathrm{e})$

$\therefore \quad \mathrm{m} _{1}=\dfrac{\mathrm{V}}{4}$ and $\left|\mathrm{C} _{1}\right|=\mathrm{e}=0.3 \mathrm{D}$

$\therefore \mathrm{V}=4 \mathrm{~m} _{1}$ and $\mathrm{D}=\dfrac{\left|\mathrm{C} _{1}\right|}{0.3} \approx \dfrac{10}{3}\left|\mathrm{C} _{1}\right|$

Also $\mathrm{m} _{2}=\dfrac{3}{4} \mathrm{~V}$ and $\left|\mathrm{C} _{2}\right|=\mathrm{e}=0.3 \mathrm{D}$

$\therefore \quad \mathrm{V}=\dfrac{4}{3} \mathrm{~m} _{2}$ and $\mathrm{D}=\dfrac{10}{3}\left|\mathrm{C} _{2}\right|$

We thus have

$\mathrm{V}=4 \mathrm{~m} _{1}=\dfrac{4}{3} \mathrm{~m} _{2}$

and

$\mathrm{D}=\dfrac{10}{3}\left|\mathrm{C} _{1}\right|=\dfrac{10}{3}\left|\mathrm{C} _{2}\right|$

Average

Specific Heat Capacity

38. A given liquid wasknown to be (chemically) non-reactive with water; a mass, $m _{L}$, of this liquid, having a (unknown) specific heat capacity, $s _{L}$, was put in a calorimeter of mass $m _{1}$, made from a material of specific heat capacity $\mathrm{s} _{\mathrm{c}}$. The liquid was heated to a temperature $\theta _{\mathrm{L}}$ and poured into another calorimeter containing water. The mass of this second calorimeter was $m _{2}$ and the mass of water, contained in it, was $m _{w}$. The specific heat capacities, of water and the material of this calorimeter, are $s _{w}$ and $s _{c}$, respectively. The initial temperature of water was $\theta _{\mathrm{C}}$. If the final temperature of the mixture was $\theta _{\mathrm{F}}$; the specific heat capacity, of the liquid, can be calculated from the formula

(1) $\mathrm{s} _{\mathrm{L}}=\dfrac{\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)-\mathrm{m} _{1} \mathrm{~s} _{\mathrm{C}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)}{\mathrm{m} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)}$

(2) $\mathrm{s} _{\mathrm{L}}=\dfrac{\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)}{\mathrm{m} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)}$

(3) $\mathrm{s} _{\mathrm{L}}=\dfrac{\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)}{\mathrm{m} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)+\mathrm{m} _{\mathrm{L}} \mathrm{s} _{\mathrm{C}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)}$

(4) $\mathrm{s} _{\mathrm{L}}=\dfrac{\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)}{\mathrm{m} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)}$

Show Answer

Correct answer: (4)

Solution:

Heat is lost only by the hot liquid. Hence heat lost $=\mathrm{m} _{\mathrm{L}} \mathrm{s} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)$

Heat is gained both by the second calorimeter and the water contained in it.

Hence heat gained $=\mathrm{m} _{\mathrm{w}} \mathrm{s} _{\mathrm{W}}\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)$

$=\left[\mathrm{m} _{\mathrm{w}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right]\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)$

Putting heat gained $=$ heat lost, we get

$\begin{aligned} & \mathrm{m} _{\mathrm{L}} \mathrm{s} _{\mathrm{L}}\left(\theta _{\mathrm{h}}-\theta _{\mathrm{F}}\right)=\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right) \\ \therefore \quad & \mathrm{s} _{\mathrm{L}}=\dfrac{\left(\mathrm{m} _{\mathrm{W}} \mathrm{s} _{\mathrm{W}}+\mathrm{m} _{2} \mathrm{~s} _{\mathrm{C}}\right)\left(\theta _{\mathrm{F}}-\theta _{\mathrm{C}}\right)}{\mathrm{m} _{\mathrm{L}}\left(\theta _{\mathrm{L}}-\theta _{\mathrm{F}}\right)} \end{aligned}$

Difficult

Specific Heat Capacity

39. When a given amount of heat (say $\mathrm{H}$ ) is supplied to a

(i) mass $m _{1}$ of a solid of specific heat capacity $s _{1}$, its rise in temperature is $\theta _{1}$.

(ii) mass $\mathbf{m} _{2}$ of a solid of specific heat capacity $\mathbf{s} _{2}$, its rise in temperature is $\theta _{2}$. Let the same amount of heat $(H)$ raise the temperature, of a mass $\left(m _{1}-m _{2}\right)$ of a substance of specific heat capacity $\left(s _{1}-s _{2}\right)$, through $\theta$. The difference, $\left[\left(\dfrac{1}{\theta _{1}}+\dfrac{1}{\theta _{2}}\right)-\dfrac{1}{\theta}\right]$, would equal

(1) zero

(2) $\dfrac{\left(\mathrm{m} _{1} \mathrm{~s} _{1}+\mathrm{m} _{2} \mathrm{~s} _{2}\right)}{\mathrm{H}}$

(3) $\dfrac{\left(\mathrm{m} _{1} \mathrm{~s} _{2}+\mathrm{m} _{2} \mathrm{~s} _{1}\right)}{\mathrm{H}}$

(4) $\dfrac{\left(\mathrm{m} _{1} \mathrm{~s} _{1}-\mathrm{m} _{2} \mathrm{~s} _{2}\right)}{\mathrm{H}}$

Show Answer

Correct answer: (2)

Solution:

We have

$\begin{aligned} \mathrm{H} & =\mathrm{m} _{1} \mathrm{~s} _{1} \theta _{1} \\ \therefore \quad \dfrac{1}{\theta _{1}} & =\dfrac{\mathrm{m} _{1} \mathrm{~s} _{1}}{\mathrm{H}} \end{aligned}$$

Similarly $\dfrac{1}{\theta _{2}}=\dfrac{\mathrm{m} _{2} \mathrm{~S} _{2}}{\mathrm{H}}$

We also have

$\begin{aligned} & \dfrac{1}{\theta}= \dfrac{\left(\mathrm{m} _{1}-\mathrm{m} _{2}\right)\left(\mathrm{s} _{1}-\mathrm{s} _{2}\right)}{\mathrm{H}} \\ &=\dfrac{\mathrm{m} _{1} \mathrm{~s} _{1}-\mathrm{m} _{2} \mathrm{~s} _{1}-\mathrm{m} _{1} \mathrm{~s} _{2}+\mathrm{m} _{2} \mathrm{~s} _{2}}{\mathrm{H}} \\ &=\dfrac{1}{\theta _{1}}+\dfrac{1}{\theta _{2}}-\dfrac{1}{\mathrm{H}}\left(\mathrm{m} _{1} \mathrm{~s} _{2}+\mathrm{m} _{2} \mathrm{~s} _{1}\right) \\ & \therefore \quad\left(\dfrac{1}{\theta _{1}}+\dfrac{1}{\theta _{2}}\right)-\dfrac{1}{\theta}=\left(\dfrac{\mathrm{m} _{1} \mathrm{~s} _{1}+\mathrm{m} _{2} \mathrm{~s} _{1}}{\mathrm{H}}\right) \end{aligned}$

Difficult

Specific Heat Capacity

40. A given solid rectangular bar $($ mass $=M)$ has an area of cross section $A$ and a length $L$. Its density $\rho$, and its specific heat capacity, $s$, vary with distance, $x$ from one end, as

$\rho(x)=\mathbf{k} _{1} x$

and $\mathrm{s}(x)=\mathbf{k} _{2} x$, respectively.

The heat needed to raise the temperature of this bar by an amount, $\theta$, is

(1) $\dfrac{1}{2} \mathrm{k} _{2} \mathrm{ML}^{3}$

(2) $\dfrac{2}{3} \mathrm{k} _{2} \mathrm{ML}$

(3) $\dfrac{1}{3} \mathrm{k} _{2} \mathrm{ML}^{3}$

(4) $\dfrac{1}{2} \mathrm{k} _{2} \mathrm{ML}^{2}$

Show Answer

Correct answer: (2)

Solution:

The mass, dm, of a segment of the bar, of length $\mathrm{d} x$, is

$\begin{aligned} & \mathrm{dm}=\rho(x) \cdot(\operatorname{Ad} x) \\ & =\mathrm{k} _{1} x \operatorname{Ad} x \end{aligned}$

The specific heat capacity, of this segment, is

$\mathrm{s}(x)=\mathrm{k} _{1} x$

Hence the heat needed to raise its temperature, by an amount $\theta$, is

$\begin{aligned} \mathrm{dH}= & \mathrm{dm} \mathrm{s}(x) \theta \\ & =\left(\mathrm{k} _{1} x \mathrm{Ad} x\right)\left(\mathrm{k} _{2} x\right) \theta \\ & =\mathrm{k} _{1} \mathrm{k} _{2} \mathrm{~A} x^{2} \theta \mathrm{d} x \end{aligned}$

Hence the total heat needed is

$\begin{gathered} \mathrm{H}=\int \limits _{0}^{\mathrm{L}} \mathrm{dH}=\mathrm{k} _{1} \mathrm{k} _{2} \mathrm{~A} \theta \int \limits _{0}^{\mathrm{L}} x^{2} \mathrm{~d} x \\ =\mathrm{k} _{1} \mathrm{k} _{2} \mathrm{~A} \theta \dfrac{\mathrm{L}^{3}}{3} \end{gathered}$

The total mass of the bar, is

$\begin{aligned} & \mathrm{M}=\int \limits _{0}^{\mathrm{L}^{\prime}} \rho(x) \mathrm{Ad} x=\mathrm{k} _{1} \mathrm{~A} \dfrac{\mathrm{L}^{2}}{2} \\ & \therefore \quad \mathrm{H}=\left(\dfrac{2}{3} \mathrm{k} _{2} \mathrm{ML} \theta\right) \end{aligned}$

Difficult

Loss of Heat by a Hot Solid

41. It is given that the rate of loss of heat by a substance, at any temperature, is proportional to the difference between that temperature and the temperature of the surroundings. A solid, of mass $80 \mathrm{~g}$, is made from a material of specific heat capacity $400 \mathrm{~J} /\left(\mathbf{k g}-{ }^{0} \mathrm{C}\right)$. It is heated to a temperature of $80^{\circ} \mathrm{C}$ and is suspended in a room at a temperature of $20^{\circ} \mathrm{C}$.

The ratio of the times, taken by this solid, in cooling (i) from $\left(70^{\circ} \mathrm{C}\right.$ to $\left.50^{\circ} \mathrm{C}\right)$ and (ii) from $\left(50^{\circ} \mathrm{C}\right.$ to $30^{\circ} \mathrm{C}$ ) would be

(1) $1: 1$

(2) $\dfrac{(\log 4)}{(\log 2)}$

(3) $\dfrac{(\log 5-\log 3)}{(\log 3)}$

(4) $\dfrac{(\log 7-\log 5)}{(\log 5-\log 3)}$

Show Answer

Correct answer: (3)

Solution:

The (small) amount of heat ( $\mathrm{dH}$ ) lost by the given solid when its temperature falls by a (small) amount $\mathrm{d} \theta$ (from $\theta$ to $(\theta-d \theta)$, is

$\mathrm{dH}=\operatorname{msd} \theta$

We are given that

$\quad-\dfrac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta _{0}\right)$

$\therefore \quad-\mathrm{ms} \dfrac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta _{0}\right)$

$\therefore \quad \dfrac{\mathrm{d} \theta}{\left(\theta-\theta _{0}\right)}=-\dfrac{\mathrm{k}}{\mathrm{ms}} \mathrm{dt}$

$\therefore \quad \int \limits _{\theta_1}^{\theta_2} \dfrac{d \theta}{\left(\theta-\theta_0\right)}=-\dfrac{k}{m s} \int \limits _0^t d t$

$\therefore \quad \ell \mathrm{n}\dfrac{\left(\theta _{1}-\theta _{0}\right)}{\left(\theta _{2}-\theta _{0}\right)}=\dfrac{\mathrm{k}}{\mathrm{ms}}(\mathrm{t})$

If follows that

(i) $\quad \ell \mathrm{n}\dfrac{(70-20)}{(50-20)}=\dfrac{\mathrm{k}}{\mathrm{ms}} \mathrm{t} _{1}$

and (ii) $\ell \mathrm{n}\dfrac{(50-20)}{(30-20)}=\dfrac{\mathrm{k}}{\mathrm{ms}} \mathrm{t} _{2}$

$\begin{aligned} \therefore \dfrac{\mathrm{t} _{1}}{\mathrm{t} _{2}} & =\left[\dfrac{\ell \mathrm{n} \left(\dfrac{5}{3}\right)}{\ell \mathrm{n} \left(\dfrac{3}{1}\right)}\right]=\dfrac{\log \left(\dfrac{5}{3}\right)}{\log (3)} \\ & =\dfrac{(\log 5-\log 3)}{(\log 3)} \end{aligned}$

Difficult

Meter Bridge

42. In a metre bridge experiment to find the resistance of a uniform wire of length $L=25 \mathrm{~cm}$, the wire is connected in the left gap. The student performing the experiment inserted $0.5 \mathrm{~cm}$ of this wire at both the ends into the terminals at the gap. The balancing length is measured to be $75 \mathrm{~cm}$, from the end nearer to the left gap. When a resistance of $2 \mathrm{ohm}$ is connected in the right gap. If the experiment is otherwise accurately performed, the actual resistance of the given wire is

(1) $6.00 \Omega$

(2) $6.25 \Omega$

(3) $5.75 \Omega$

(4) $6.40 \Omega$

Show Answer

Correct answer: (2)

Solution:

Let $\mathrm{X}$ ’ be the resistance of the ’effective’ length of the wire.

The effective length is $\mathrm{L}^{\prime}=(\mathrm{L}-0.5-0.5 \mathrm{~cm})=24 \mathrm{~cm}$ i.e. $\mathrm{X}^{\prime}$ is resistance of $24 \mathrm{~cm}$ of the given wire. Now, $\mathrm{R}=2 \Omega$ and balancing length is $\ell=75 \mathrm{~cm}$. As the balancing length is measured from the end where the wire is connected, the relevant formla is:

$X^{\prime}=\left(\dfrac{\ell}{100-\ell}\right) R=\left(\dfrac{75}{25} \times 2\right) \Omega=6 \Omega$

Now, $\mathrm{b} \Omega$ is the resistance of $24 \mathrm{~cm}$ of the wire [(excluding) the length inserted in to the terminal, which does not provide any resistance)].

$\therefore$ The actual resistance $\mathrm{X}$ is:

$X=\left(\dfrac{25}{24} \times 6\right) \Omega=6.25 \Omega$

Average

Meter Bridge

43. A metre bridge circuit, with resistance $X$ and $R$ connected in the gaps, gives a balance length of $\ell$, from the end where $X$ is connected. When a resistance of $5 \mathrm{P}$ is connected in parallel with $X$ and another resistance of $15 Q$ is connected in parallel with $R$, the balancing length remained the same. The raio of $P$ to $Q$, in terms of $\ell$, is

(1) $\dfrac{3 \ell}{(100-3 \ell)}$

(2) $\dfrac{\ell}{(100-3 \ell)}$

(3) $\dfrac{\ell}{(300-\ell)}$

(4) $\dfrac{3 \ell}{(100-\ell)}$

Show Answer

Correct answer: (4)

Solution:

Given $\dfrac{X}{R}=\dfrac{\ell}{100-\ell}$

When $5 \mathrm{P}$ is connected is parallel with $\mathrm{X}$, the effective resistance in Gap 1 is $\dfrac{\mathrm{X} \times 5 \mathrm{P}}{\mathrm{X}+5 \mathrm{P}}$

When $15 \mathrm{Q}$ is connected in parallel with $\mathrm{R}$, the effective resistance in Gap 2 is $\dfrac{R \times 15 \mathrm{Q}}{\mathrm{R}+15 \mathrm{Q}}$

Since balancing length remains same,

$\dfrac{\mathrm{X} \times 5 \mathrm{P}}{(\mathrm{X}+5 \mathrm{P})} \times \dfrac{(\mathrm{R}+15 \mathrm{Q})}{\mathrm{R} \times 15 \mathrm{Q}}=\dfrac{\mathrm{X}}{\mathrm{R}}$

or $75 \mathrm{PQ}+5 \mathrm{PR}=15 \mathrm{QX}+75 \mathrm{PQ}$

or $\dfrac{\mathrm{P}}{3 \mathrm{Q}}=\dfrac{\mathrm{X}}{\mathrm{R}}$

or $\dfrac{\mathrm{P}}{\mathrm{Q}}=3\left(\dfrac{\mathrm{X}}{\mathrm{R}}\right)$

or $\dfrac{\mathrm{P}}{\mathrm{Q}}=3\left(\dfrac{\ell}{100-\ell}\right)$

Alternatively: Since the balacing length is the same, potential drops across the gaps remain the same. Therefore, the ratio of resistances would be the same along individual paths.

$\therefore \dfrac{\mathrm{X}}{\mathrm{R}}=\dfrac{3 \mathrm{P}}{15 \mathrm{Q}}$

or $\dfrac{P}{Q}=3 \dfrac{X}{R}=\dfrac{3 \ell}{100-\ell}$

Easy

Meter Bridge

44. In a metre bridge experiment, to determine the specific resistance ( $\rho$ ) of the material of the given wire, the percentage error in the measurement of resistance $X$, diameter (d) and length ( $L$ ) of the wire are respectively $1.25 \%, 0.75 \%$ and $0.25 \%$ respectively. The percentage error in the measurement of $\rho$ is:

(1) $1.75 \%$

(2) $2.25 \%$

(3) $2.50 \%$

(4) $3.00 \%$

Show Answer

Correct answer: (4)

Solution:

The formula for $\rho$, in the case of metre bridge experiment, is:

$\begin{aligned} \rho & =\dfrac{\mathrm{X} \dfrac{\pi \mathrm{d}^{2}}{4}}{\mathrm{~L}} \\ \therefore \quad \dfrac{\Delta \rho}{\rho} & =\dfrac{\Delta \mathrm{X}}{\mathrm{X}}+2 \dfrac{\Delta \mathrm{d}}{\mathrm{d}}+\dfrac{\Delta \ell}{\mathrm{L}} \\ & =1.25 \%+2 \times 0.75 \%+0.25 \% \\ & =1.25 \%+1.5 \%+0.25 \% \\ & =3.00 \% \end{aligned} $

Difficult

Ohm’s Law

45. While performing the experiment, to find the resistance $(R)$ of a wire, using a d.c source of 10 volt, a miliammeter of negligible resistance, a voltmeter and a rheostat of total resistance $200 \Omega$, a student connected all, including the voltmeter, in series, by mistake. The sliding contact of the rheostat was kept exactly at the middle, as shown in the circuit diagram.

If the milliammeter read $3.2 \mathrm{~mA}$ and the voltmeter read $9.6 \mathrm{~V}$, the resistance of the voltmeter and the value of $R$ are, respectively,

(1) $3125 \Omega$ and $125 \Omega$

(2) $3000 \Omega$ and $25 \Omega$

(3) $3125 \Omega$ and $25 \Omega$

(4) $3100 \Omega$ and $125 \Omega$

Show Answer

Correct answer: (2)

Solution:

In series also, a voltmeter will still read the potential difference across its terminals.

$\therefore$ Resistance of voltmeter $=\dfrac{\text { Reading in volt }}{\text { current in ampere }}$

or $\mathrm{R} _{\mathrm{V}}=\dfrac{9.6}{3.2 \times 10^{-3}}=3000 \mathrm{~V}$

The resistance, due to the rheostat, is half of its total resistance of $200 \Omega$

$\therefore$ The potential drop over the rheostat $=3.2 \mathrm{~mA} \times 100 \Omega$

$=0.32$ volt

Total potential applied $=10 \mathrm{~V}$

$\therefore$ The potential drop across $\mathrm{R}=10-(9.6+0.32)$

$=0.08$ volt

Therefore, the value of $\mathrm{R}=\dfrac{\text { Voltage across } \mathrm{R}}{\text { Current }}$

$\begin{aligned} & =\dfrac{0.08}{3.2 \times 10^{-3}} \\ & =25 \Omega \end{aligned}$

Average

Ohm’s Law

46. While performing the experiment on Ohm’s law, a student plotted the V-I graphs for two resistances, individually and for their series and parallel combinations. The four plots are marked A, B, C and D in the graph shown below. The values of the individual resistances in ohm, are approximately

(1) 1 and $\dfrac{1}{\sqrt{3}}$

(2) 1 and $\sqrt{3}$

(3) 1 and 0.732

(4) $\sqrt{3}$ and $\dfrac{1}{\sqrt{3}}$

Show Answer

Correct answer: (3)

Solution:

Since $R=\dfrac{V}{I}$, the reciprocal of the slope of the plots will give the resistance. Therefore, the resistance of

A is maximum and the value of $\mathrm{R} _{\mathrm{A}}=\dfrac{1}{\tan 30^{\circ}}=\sqrt{3}=1.732 \Omega$

This must be the resistance for the series combination. Since the resistance is the least for $\mathrm{D}$, this must be for the parallel combination.

Therefore graphs B and C are for the individual resistances.

B graph representing one of the resistances, has its slope as unity. Hence this resistance is $1 \Omega$.

$\therefore \quad \mathrm{R} _{\mathrm{A}}=\mathrm{R} _{\text {series }}=\sqrt{3}=\mathrm{R} _{\mathrm{B}}+\mathrm{R} _{\mathrm{C}}=1+\mathrm{R} _{\mathrm{C}}$

$\therefore \mathrm{R} _{\mathrm{C}}=\sqrt{3}-1=0.732 \Omega$

Hence the two resistances are $1 \Omega$ and $0.732 \Omega$.

Average

Ohm’s Law

47. Using the Ohm’s circuit, for determination of resistnace of a wire made of an alloy of constant temperature coefficient of resistance $(\alpha)$, a student obtained the resistance of the wire at $0^{\circ} \mathrm{C}$, at $100^{\circ} \mathrm{C}$ and at $t^{\circ} \mathrm{C}$ as $3.14 \Omega, 3.54 \Omega$ and $3.42 \Omega$ respectively. The temperature $t$ is

(1) $60^{\circ} \mathrm{C}$

(2) $70^{\circ} \mathrm{C}$

(3) $80^{\circ} \mathrm{C}$

(4) $90^{\circ} \mathrm{C}$

Show Answer

Correct answer: (2)

Solution:

The variation of $\mathrm{R}$, with temperature, is expressed as

$\mathrm{R} _{(\mathrm{t})}=\mathrm{R} _{(0)}(1+\alpha \mathrm{t})$

Hence we have

$\begin{aligned} & \mathrm{R} _{(100)}=\mathrm{R} _{(0)}(1+\alpha \times 100) \\ & \mathrm{R} _{(\mathrm{t})}=\mathrm{R} _{(0)}(1+\alpha \mathrm{t}) \\ \therefore \quad \mathrm{t} & =\left(\dfrac{\mathrm{R} _{(\mathrm{t})}-\mathrm{R} _{(0)}}{\mathrm{R} _{(100)}-\mathrm{R} _{(0)}}\right) \times 100 \\ & =\left(\dfrac{3.42-3.14}{3.54-3.14}\right) \times 100 \\ & =\dfrac{0.28}{0.40} \times 100=70^{\circ} \mathrm{C} \end{aligned}$

Easy

Potentiometer

48. Which of the following statements is incorrect regarding the use of the high resistance in the potentiometer circuit?

(1) It protects the galvanometer from being damaged due to high current.

(2) Its value does not play any role in the determination of null point.

(3) It must always be necessarily connected just before the galvanometer to prevent high current entering the galvanometer.

(4) It can be short circuited when the jockey, moved along the wire, approaches the null point position.

Show Answer

Correct answer: (3)

Solution:

(1) The galvanometer circuit can have large current when the jockey is pressed near the extreme ends of the poteniomeer wire due to the large potential difference across the wire, far away from the null point. Since the galvanometer is highly sensitive, it might get damaged due to this high current. Ahigh resistance will reduce this current and protect the galvanometer. Hence (1) is a correct statement.

(2) At the null point the deflection in the galvanometer is zero as the current is zero. So the value of this high resistance does not decide the position of the null point, it only decreases the deflection when jockey is pressured at other points. So (2) is also a correct statement.

(3) Since a resistor only obstruct the flow of charge, the position of the resistor can be anywhere in the series part of the circuit. That means, the high resistance can be before, or after, the galvanometer. In either case the current gets reduced. So (3) is an incorrect statement (i.e. (3) is the correct answer to the question).

(4) When we approach the null point, the current through the galvanometer approaches zero or the deflection becomes almost zero. Here if we short circuit the high resistance, the deflection becomes obviously more, but the current will not to so high as to damage the galvanometer. Instead the short circuit of the high resistance will enable us to find the null point accurately. So (4) is also a correct statement.

Average

Potentiometer Internal Resistance

49. In a potentiometer experiment the following circuit arrangement is used. $A B$ is the potentiometer wire.

When the key $K _{1}$ is closed ( $K _{2}$ open), the balancing length is found to be $\ell$. When the key $K _{2}$ is closed ( $K _{1}$ open), the balancing length is found to be $2 \ell$. The value of $r$, in terms of $R _{1}$ and $R _{2}$, is:

(1) $\dfrac{R _{1} R _{2}}{R _{2}-R _{1}}$

(2) $\dfrac{2 R _{1} R _{2}}{R _{2}-R _{1}}$

(3) $\dfrac{R _{1} R _{2}}{R _{2}-2 R _{1}}$

(4) $\dfrac{2 R _{1} R _{2}}{R _{2}-2 R _{1}}$

Show Answer

Correct answer: (3)

Solution:

Let $\mathrm{V} _{1}$ be the terminal voltage when $\mathrm{K} _{1}$ is closed, that is, when $\mathrm{R} _{1}$ is connected across the cell of emf $\mathrm{E}$ and internal resistance $\mathrm{r}$.

Here $V _{1}=\dfrac{E R _{1}}{R _{1}+r}$

But $\mathrm{V} _{1}=\mathrm{k} \ell \quad\left(\ell=\right.$ balancing length when $\mathrm{K} _{1}$ is closed (given)

$\therefore \quad \mathrm{k} \ell=\dfrac{\mathrm{ER} _{1}}{\mathrm{R} _{1}+\mathrm{r}}$ $\hspace {3 cm}$ …….(1)

Let $\mathrm{V} _{2}$ be the terminal voltage when $\mathrm{K} _{2}$ is closed.

Then $V _{2}=\dfrac{E R _{2}}{R _{2}+r}$

But $\mathrm{V} _{2}=\mathrm{k}(2 \ell) \quad(2 \ell=$ balacing length $)$

$\therefore \mathrm{k}(2 \ell)=\dfrac{\mathrm{ER} _{2}}{\mathrm{R} _{2}+\mathrm{r}}$ $\hspace {3 cm}$ …….(2)

Dividing (1) by (2), we get:

$\dfrac{1}{2}=\dfrac{\mathrm{R} _{1}\left(\mathrm{R} _{2}+\mathrm{r}\right)}{\mathrm{R} _{2}\left(\mathrm{R} _{1}+\mathrm{r}\right)}$

or $R _{2} R _{1}+R _{2} r=2 R _{1} R _{2}+2 R _{1} r$

or $\quad\left(R _{2}-2 R _{1}\right) r=R _{1} R _{2}$

$\therefore \quad \mathrm{r}=\dfrac{\mathrm{R} _{1} \mathrm{R} _{2}}{\mathrm{R} _{2}-2 \mathrm{R} _{1}}$

Difficult

Potentiometer - Comparison of emfs

50. A potentiometer wire $A B$ is of length $100 \mathrm{~cm}$, with $C$ as its middle point. The part $A C$ is made of a uniform wire of a material of resistivity $\rho _{1}$. The diameter of this part is $d _{1}$. The part CB is made of another uniform wire of a material of resistivity $\rho _{2}$. The diameter of this part $d _{2}$. It is given that

$\rho _{1}: \rho _{2}=2: 3 \text { and } d _{1}: d _{2}=2: 1$

The balacing length, measured from $A$, for a cell for $E M F E _{1}$ is found to be $40 \mathrm{~cm}$ and that for a cell of $E M F E _{2}$, is found to be $65 \mathrm{~cm}$. The ratio $E _{1} / E _{2}$ is equal to

(1) $\dfrac{2}{7}$

(2) $\dfrac{4}{5}$

(3) $\dfrac{40}{65}$

(4) $\dfrac{4}{9}$

Show Answer

Correct answer: (1)

Solution:

Given $\dfrac{\rho _{1}}{\rho _{2}}=\dfrac{2}{3} \quad \therefore \rho _{2}=\dfrac{3}{2} \rho _{1}$

$\dfrac{\mathrm{d} _{1}}{\mathrm{~d} _{2}}=\dfrac{2}{1} \quad \therefore \quad \mathrm{d} _{2}=\dfrac{1}{2} \mathrm{~d} _{1}$

Now, let $\sigma _{1}$ and $\sigma _{2}$ be the resistance per unit length of each half.

$\sigma _{1}=\dfrac{\rho _{1} \times 1}{\pi \dfrac{\mathrm{d} _{1}^{2}}{4}} \text { and } \sigma _{2}=\dfrac{\rho _{2} \times 1}{\pi \dfrac{\mathrm{d} _{2}^{2}}{4}}=\dfrac{\dfrac{3}{2} \rho _{1}}{\dfrac{\pi}{4}\left(\dfrac{\mathrm{d} _{1}}{2}\right)^{2}}=\dfrac{\rho _{1}}{\pi \dfrac{\mathrm{d} _{1}^{2}}{4}} \times 6$

$\therefore \quad \sigma _{2}=6 \sigma _{1}$

Let I be the current through the wire, which must be the same for both the halves.

By using the principle of potentiometer, $\mathrm{E}=\mathrm{I} \sigma \ell$

$\mathrm{E} _{1}=\mathrm{I} \sigma _{1} \ell _{1}=\mathrm{I} \sigma _{1} \times 40$

and

$\mathrm{E} _{2}=\mathrm{I}\left(\sigma _{1} \times 50+\sigma _{2} \times 15\right)$ $\hspace {2 cm}$ $\left(\because \ell _{2}=65 \mathrm{~cm}=50 \mathrm{~cm}+15 \mathrm{~cm}\right)$

$\therefore \quad \dfrac{\mathrm{E} _{1}}{\mathrm{E} _{2}}=\dfrac{\sigma _{1} \times 40}{\sigma _{1} \times 50+6 \sigma _{1} \times 15}$

$=\dfrac{4}{5+9}=\dfrac{4}{14}=\dfrac{2}{7}$

Difficult

Potentiometer Application

51. For the potentiomeer circuit, set up as shown, the potential gradient, across the potentiometer wire, $A B$, is (k) volt $/ \mathrm{cm}$. The balancing lengths, with the gap between (i) points (1) and (3) plugged in, equal $\left(\ell _{1}\right) \mathrm{cm}$ and (ii) points (2) and (3) plugged in, equals $\left(\ell _{2}\right) \mathrm{cm}$.

Assuming that the current value, as obtained through the potentiometer readings, is the correct value of current, the percentage error, in the reading of the ammeter, equals

(1) $\dfrac{100}{\mathrm{k} \ell _{1}}\left(\mathrm{I} _{0} \mathrm{R} _{1}-\mathrm{k} \ell _{1}\right)$ or $\dfrac{\left(\mathrm{I} _{0}\left(\mathrm{R} _{1}+\mathrm{R} _{2}\right)-\mathrm{k} \ell _{2}\right) 100}{\left(\mathrm{k} \ell _{2}\right)}$

(2) $\dfrac{100}{\mathrm{k} \ell _{2}}\left(\mathrm{I} _{0} \mathrm{R} _{1}-\mathrm{k} \ell _{1}\right)$ or $\dfrac{100}{\mathrm{k} \ell _{1}}\left[\mathrm{I} _{0}\left(\mathrm{R} _{1}+\mathrm{R} _{2}\right)-\mathrm{k} \ell _{2}\right]$

(3) $\dfrac{100}{\mathrm{k} \ell _{1}}\left(\mathrm{I} _{0}\left(\mathrm{R} _{1}-\mathrm{R} _{2}\right)-\mathrm{k} \ell _{1}\right)$ or $\dfrac{100}{\mathrm{k} \ell _{2}}\left[\mathrm{I} _{0} \mathrm{R} _{1}-\mathrm{k} \ell _{2}\right]$

(4) $\dfrac{100}{\mathrm{k} \ell _{2}}\left(\mathrm{I} _{0} \mathrm{R} _{2}-\mathrm{k} \ell _{2}\right)$ or $\dfrac{100}{\mathrm{k} \ell _{1}}\left[\mathrm{I} _{0}\left(\mathrm{R} _{1}-\mathrm{R} _{2}\right)-\mathrm{k} \ell _{1}\right]$

Show Answer

Correct answer: (1)

Solution:

When the gap between points (1) and (3) is plugged in, we have

Potential drop across $\mathrm{R} _{1}=\mathrm{k} \ell _{1}$

$\therefore$ Current through $\mathrm{R} _{1}=\dfrac{\mathrm{k} \ell _{1}}{\mathrm{R} _{1}}$

Current read by the ammeter $=\mathrm{I} _{0}$

$\therefore$ Error in ammeter reading $=\left(\mathrm{I} _{0}-\dfrac{\mathrm{k} \ell _{1}}{\mathrm{R} _{1}}\right)$

$\therefore$ Percentage error $=\left[\dfrac{\mathrm{I} _{0} \mathrm{R} _{1}-\mathrm{k} \ell _{1}}{\mathrm{R} _{1}} \times \dfrac{\mathrm{R} _{1}}{\mathrm{k} \ell _{1}}\right] \times 100$

$=\left(\mathrm{I} _{0} \mathrm{R} _{1}-\mathrm{k} \ell _{1}\right) \dfrac{100}{\mathrm{k} \ell _{1}}$ $\hspace{3 cm}$ ……..(1)

Similarly, with the gap between points (2) and (3) plugged in, we get the percentage error in ammeter reading as

$\left(\mathrm{I} _{0}\left(\mathrm{R} _{1}+\mathrm{R} _{2}\right)-\mathrm{k} \ell _{2}\right)\left(\dfrac{100}{\mathrm{k} \ell _{2}}\right)$ $\hspace{3 cm}$ ……..(2)

The percentage error can thus have either of the values given by expressions (1) and (2).

Difficult

Galvanometer Resistance

52. A galvanometer, of unknown figure of merit $(k)$ and unknown resistance $(G)$, is connected in series with a cell of emf $E$ and a variable resistance $R$. The deflection, obtained in the galvanometer, is noted as $\theta$ divisions for every resistance value.

The experimental values of $R$ and $\theta$ are used to plot a graph of $\left(\dfrac{1}{\theta}\right) \operatorname{div}^{-1}$ along the $y$-axis and $(R) \Omega$ along the the $x$-axis. From the graph, the resistance $G$ of the galvanometer can be obtained as the

(1) Slope of the graph

(2) $y$-intercept of the graph

(3) Ratio of y-intercept to the slope of the graph

(4) Product of y-intercept and the slope of the graph

Show Answer

Correct answer: (3)

Solution:

From the circuit, the value of I can be written as:

$I=\dfrac{E}{R+G}$

or $\mathrm{K} \theta=\dfrac{\mathrm{E}}{\mathrm{R}+\mathrm{G}}$

or $\dfrac{1}{\theta}=\left(\dfrac{K}{E}\right) R+\left(\dfrac{K}{E}\right) \mathrm{G}$

Since $\mathrm{R}$ and $\theta$ are the only variables here, the above equation is of the form

$\mathrm{y}=\mathrm{m} x+\mathrm{C}$, which gives a straight line. Here $\mathrm{m}=\dfrac{\mathrm{K}}{\mathrm{E}}$ is the slope of the straight line graph and $C=\left(\dfrac{K}{E}\right) G$ is the $y-$ intercept.

Now, $\dfrac{\mathrm{C}}{\mathrm{m}}=\dfrac{\left(\dfrac{\mathrm{K}}{\mathrm{E}}\right) \mathrm{G}}{\left(\dfrac{\mathrm{K}}{\mathrm{E}}\right)}=\mathrm{G}$

$\therefore G=$ ratio of $y$-intercept to the slope of the graph.

Easy

Galvanometer – G and K

53. In the galvanometer experiment, the values of $G$ and $k$ are obtained as $300 \Omega$ and $1.5 \times 10^{-4}$ ampere/division, for a galvanometer having a total 30 divisions. A student wants to use this galvanometer as a voltmeter to measure a voltage up to 9 volt, for his project. He noticed that carbon resistors of the following sequence of colour bands, are available in the laboratory.

$$ \begin{aligned} & \mathbf{R} _{1} \rightarrow \text { Brown, Blue, } \text { Red } \\ & \mathbf{R} _{2} \rightarrow \text { Black, Violet, } \text { Red } \\ & \mathbf{R} _{3} \rightarrow \text { Brown, Yellow, } \text { Red } \\ & \mathbf{R} _{4} \rightarrow \text { Brown, Violet, Red } \end{aligned} $$

He would use the resistor:

(1) $\mathrm{R} _{1}$

(2) $\mathrm{R} _{2}$

(3) $\mathrm{R} _{3}$

(4) $\mathrm{R} _{4}$

Show Answer

Correct answer: (4)

Solution:

To convert the galvanometer in to a voltmeter, he has to connect a series resistor $\mathrm{R}$ with the galvanometer such that when $\mathrm{V}=9$ volt, that voltmeter will give full-scale deflection, or a current $\mathrm{i} _{\mathrm{g}}$ flows through the galvanometer (when 9 volt is applied).

The value of $R$ is:

$\mathrm{R}=\dfrac{\mathrm{V}}{\mathrm{I} _{\mathrm{g}}}-\mathrm{G}$

Here, $\mathrm{V}=9$ volt

$I _{\mathrm{g}}=30 \times \mathrm{k}$

$=30 \times 1.5 \times 10^{-4} \mathrm{~A}$

$=4.5 \times 10^{-3} \mathrm{~A}$

$\mathrm{G}=300 \Omega$

$\begin{aligned} \therefore \quad \mathrm{R} & =\dfrac{9}{4.5 \times 10^{-3}}-300 \\ & =1700 \Omega \end{aligned}$

The colour code corresponding to this $\mathrm{R}$ is:

Brown, Violet, Red.

Therefore $\mathrm{R} _{4}$ is to be used as $\mathrm{R}$.

Easy

Half Deflection Method

54. While performing the galvanometer experiment using half deflection method, a student got almost the same value for $S$, every time, whatever be the value of $R$ he had set in the (high value) resistance box. This is because

(1) The resistance box in which $\mathrm{S}$ is put is faulty

(2) Resistance G is very low

(3) In half-deflection method, current gets divided between $\mathrm{G}$ and $\mathrm{S}$ and $\mathrm{R}$ is very high

(4) The internal resistance of the d.c source varies with current flow

Show Answer

Correct answer: (3)

Solution:

Here, the galvanometer is a very sensitive device. The current flowing in it should be very low. So $\mathrm{R}$ should be quite high.

The formula for $\mathrm{G}$ is

$\mathrm{G}=\dfrac{\mathrm{RS}}{\mathrm{R}-\mathrm{S}}$

Since $\mathrm{R}$ is large, $(\mathrm{R}-\mathrm{S})$ is almost equal to $\mathrm{R}$.

This shows that, $\mathrm{G} \approx \dfrac{\mathrm{RS}}{\mathrm{R}} \approx \mathrm{S}$

Difficult

Finding the Resistance of Galvanometer

55. For the circuit shown here, the galvanometer deflection equals (i) 15 divisons when the key $K _{2}$ is kept open and (ii) 3 divisions when key $K _{2}$ is closed. The resistance of this galvanometer is (nearly)

(1) $60 \Omega$

(2) $64 \Omega$

(3) $68 \Omega$

(4) $75 \Omega$

Show Answer

Correct answer: (2)

Solution:

With the key $\mathrm{K} _{2}$ open, the current, through the galvanometer, is

$i _{1}=\dfrac{\varepsilon}{R+G}$

When the key $K _{2}$ is closed, the total resistance of the circuit equals $\left[R+\dfrac{G S}{(G+S)}\right]$. Hence the current, through the galvanometer, is

$i _{2}=\dfrac{\varepsilon}{\left[R+\left(\dfrac{G S}{G+S}\right)\right]} \times\left(\dfrac{S}{G+S}\right)=\dfrac{\varepsilon S}{R G+G S+S R}$

We are given that $\mathrm{i} _{2}=\dfrac{3}{15} \mathrm{i} _{1}=\dfrac{\mathrm{i} _{1}}{5}$

$\therefore 5 \mathrm{RS}+5 \mathrm{GS}=\mathrm{RG}+\mathrm{GS}+\mathrm{SR}$

$\therefore \mathrm{G}[4 \mathrm{~S}-\mathrm{R}]=-4 \mathrm{RS}$

$\therefore \mathrm{G}=\dfrac{4 \mathrm{RS}}{\mathrm{R}-4 \mathrm{~S}}$

$\begin{aligned} & =\dfrac{4 \times 1000 \times 15}{(1000-60)} \Omega=\left(\dfrac{1000}{940} \times 60\right) \Omega \\ & =63.8 \Omega \simeq 64 \Omega \end{aligned}$

Average

Focal Length of Concave Mirror

56. In the experiment to find the focal length of a concave mirror, using parallex method, a student placed the object pin at a point so at to obtain an enlarged real image. He used an image pin in order to locate the position of this real image of the object pin, by trying to remove the parallex between the two. Checking for parallex, he observed that the image of the object pin moved through greater distance than the image pin, with respect to the edge of the mirror, when he moved his eye. The position, across the optical bench (with respect to the present position of the image pin), of the image (of the object-pin) is:

(1) Closer to his eyes.

(2) Closer to the centre of curvature of the mirror.

(3) between the image pin and the centre of curvature of the mirror.

(4) At the position of the object pin.

Show Answer

Correct answer: (1)

Solution:

It is obvious that more the relative distance moved closer will be the object observed (here the image of the object-pin), with respect to the other.Hence the image (of the object pin) is closer to his eyes.

Average

Focal Length of Convex Lens

57. While using the displacement method, for determination of focal length of a convex lens, a student placed the object pin at $10 \mathrm{~cm}$ mark on the optical bench and the imge pin at $90 \mathrm{~cm}$ mark. He was able to obtain the image of the object pin, over the image pin, without any parallax for two different positions of the convex lens. He observed that, in order to obtain these two positions of the lens, he had to move the lens from the first position to the second position through $40 \mathrm{~cm}$. The focal length of the lens is:

(1) $10 \mathrm{~cm}$

(2) $15 \mathrm{~cm}$

(3) $20 \mathrm{~cm}$

(4) $40 \mathrm{~cm}$

Show Answer

Correct answer: (2)

Solution:

The formula for $\mathrm{f}$ in the displacement method is:

$\begin{aligned} & \mathrm{f}=\dfrac{\left(\mathrm{D}^{2}-x^{2}\right)}{4 \mathrm{D}} \operatorname{Here}(\mathrm{D}=(90-10) \mathrm{cm}=80 \mathrm{~cm}, x=40 \mathrm{~cm}) \\ & \therefore \quad \mathrm{f}=\dfrac{\left(80^{2}-40^{2}\right)}{4 \times 80} \\ & \quad=\dfrac{6400-1600}{320} \mathrm{~cm} \\ & \quad=\dfrac{4800}{320} \mathrm{~cm}=15 \mathrm{~cm} \end{aligned}$

Average

Focal Length of Convex Lens

58. In the experiment to determine the focal length of a convex lens, using the method of parallex, the following observation were recorded by a student. Which of the observations is possibly not taken correctly by the student?

Sl. No.	Position of the object $(\mathrm{cm})$	Position of Lens $(\mathrm{cm})$	Position of Image $(\mathrm{cm})$
1.	25	50	66.7
2.	30	50	70.0
3.	35	50	80.0
4.	38	50	88.0
5.	20	50	65.0
6.	15	50	64.0

(1) S. No. 6

(2) S. No. 5

(3) S. No. 4

(4) S. No. 3

Show Answer

Correct answer: (3)

Solution:

Obtaining the values of $u$ and $v$ for each set and calculating the value fusing the formula $\dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}-\dfrac{1}{\mathrm{u}}$, we get the following table.

SI. No.	Position of the object $(\mathrm{cm})$	Position of Lens $(\mathrm{cm})$	Position of Imagc $(\mathrm{cm})$	$(-) \mathrm{u}$ $(\mathrm{cm})$	$\mathrm{V}$ $(\mathrm{cm})$	$\mathrm{f}=\dfrac{\mathrm{u} \times \mathrm{v}}{\mathrm{u}-\mathrm{v}}$ $(\mathrm{cmI})$
1.	25	50	66.7	25	16.7	10.0
2.	30	50	70.0	20	20.0	10.0
3.	35	50	80.0	15	30.0	10.0
4.	38	50	88.0	12	38.0	9.1
5.	20	50	65.0	30	15.0	10.0
6.	15	50	64.0	35	14.0	10.0

Clearly, observation at Sl. No. 4, has not been taken correctly.

Average

Focal Length of Convex Lens

59. A convex lens, of focal length, $f _{1}$, and a convex mirror (having its centre of curvature at the point $C$ ), of focal length, $f _{2}$, are set up as shown. It is observed that, for this set up, there is no parallex between an object, put at $A$, and its image, formed by the ‘combination’. The ratio of the focal length, $f _{1}$ and $f _{2}$, is then equal, to

(1) $\dfrac{\mathrm{d}(\mathrm{a}+x+\mathrm{d})}{2 \mathrm{a}(x+\mathrm{d})}$

(2) $\dfrac{\mathrm{a}(\mathrm{a}+x+\mathrm{d})}{2 \mathrm{a}(x+\mathrm{d})}$

(3) $\dfrac{2 \mathrm{a}(x+\mathrm{d})}{\mathrm{d}(\mathrm{a}+x+\mathrm{d})}$

(4) $\dfrac{2 \mathrm{d}(x+\mathrm{d})}{\mathrm{a}(\mathrm{a}+x+\mathrm{d})}$

Show Answer

Correct answer: (3)

Solution:

The image of the object, put at A, coincides with the object itself. This implies that the rays, redfracted by the convex lens, are incident normally on the convex mirror. Hence the image, formed by the convex lens alone, is at a distance $(x+\mathrm{d})$, from the lens. For the lens, we then have

$\begin{aligned} & v=(x+\mathrm{d}) \text { and } \mathrm{u}=-\mathrm{a} \\ \therefore \quad & \dfrac{1}{x+\mathrm{d}}-\dfrac{1}{(-\mathrm{a})}=\dfrac{1}{\mathrm{f} _{1}} \end{aligned}$

This gives $\mathrm{f} _{1}=\dfrac{\mathrm{a}(x+\mathrm{d})}{(\mathrm{a}+x+\mathrm{d})}$

We are given that $\mathrm{C}$ is the centre of curvature of the mirror.

Hence $\mathrm{f} _{2}=\dfrac{\mathrm{d}}{2}$

$\therefore \dfrac{\mathrm{f} _{1}}{\mathrm{f} _{2}}=\dfrac{2 \mathrm{a}(x+\mathrm{d})}{\mathrm{d}(\mathrm{a}+x+\mathrm{d})}$

Average

Redfraction of Lightly Prism

60. A ray of light is incident at grazing incidence on one face of an equilateral prism made from a transparent material of redfractive index $\sqrt{2}$. The angle of emergence, and the angle of deviation, for this ray, would be equal to (respectively),

(1) $\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right] \text { and }\left\{\dfrac{\pi}{6}+\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right]\right\}$

(2) $\sin ^{-1}\left[\sqrt{2} \sin 75^{0}\right]$ and $\left\{\dfrac{\pi}{6}+\sin ^{-1}\left[\sqrt{2} \cos 15^{0}\right]\right\}$

(3) $\sin ^{-1}\left[\sqrt{2} \sin 75^{0}\right] \text { and }\left\{\dfrac{\pi}{6}+\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right]\right\}$

(4) $\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right] \text { and }\left\{\dfrac{\pi}{6}+\sin ^{-1}\left[\sqrt{2} \cos 15^{0}\right]\right\}$

Show Answer

Correct answer: (1)

Solution:

For the ray, incident at grazing incidence, on one face, the corresponding angle of redfraction, $\mathrm{r} _{1}$, would be given by

$\dfrac{\sin 90^{\circ}}{\sin r _{1}}=\mu$

$\therefore \sin \mathrm{r} _{1}=\dfrac{1}{\mu}=\dfrac{1}{\sqrt{2}}$

$\therefore \mathrm{r} _{1}=45^{\circ}$

If angle of incidence, at the second face, is $r _{2}$, we would have

$\begin{array}{ll} & \mathrm{r} _{1}+\mathrm{r} _{2}=\mathrm{A}=60^{0} \\ \therefore & \mathrm{r} _{2}=15^{\circ} \end{array}$

$\therefore$ If e is the angle of emergence, we have

$\dfrac{\sin \mathrm{e}}{\sin 15^{\circ}}=\mu=\sqrt{2}$

$\therefore \quad \mathrm{e}=\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right]$

Also, if $\mathrm{D}$ is the angle of deviation, we have

$\begin{aligned} \therefore \quad & \mathrm{i}+\mathrm{e}=\mathrm{A}+\mathrm{D} \\ & \mathrm{D}=\mathrm{i}+\mathrm{e}-\mathrm{A} \\ & =90^{0}+\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right]-60^{\circ} \\ & =30^{0}+\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right] \\ & =\left\{\dfrac{\pi}{6}+\sin ^{-1}\left[\sqrt{2} \sin 15^{0}\right]\right\} \end{aligned}$

Difficult

Refractive of Light Through a Prism

61. A ray of light, incident on the face $A B$ of a prism (made from a material of redfractive index $\mu$ ) gets redfracted there and falls on its other face $A C$. If it were to go grazing along this face, the angle of redfraction $(<r)$, at the face $A B$, would equal

(1) $\left(\mathrm{A}+\sin ^{-1} \dfrac{1}{\mu}\right)$ as well as $\sin ^{-1}\left[\sin \left\{\cos ^{-1}(\sin (\mathrm{A}+\mathrm{D}))\right\}\right]$

(2) $\left(\mathrm{A}+\sin ^{-1} \dfrac{1}{\mu}\right)$ as well as $\sin ^{-1}\left[\cos \left\{\sin ^{-1}(\cos (\mathrm{A}+\mathrm{D}))\right\}\right]$

(3) $\left(\mathrm{A}-\sin ^{-1} \dfrac{1}{\mu}\right)$ as well as $\sin ^{-1}\left[\cos \left\{\sin ^{-1}(\cos (\mathrm{A}+\mathrm{D}))\right\}\right]$

(4) $\left(\mathrm{A}-\sin ^{-1} \dfrac{1}{\mu}\right)$ as well as $\sin ^{-1}\left[\sin \left\{\cos ^{-1}(\sin (\mathrm{A}+\mathrm{D}))\right\}\right]$

Show Answer

Correct answer: (4)

Solution:

It is clear that the angle of emergence, at the face $\mathrm{AC}$, is $90^{\circ}$. The angle of incidence (say $\mathrm{r}^{\prime}$ ), at the face $\mathrm{AC}$, must equal the critical angle $\left(=\sin ^{-1} \dfrac{1}{\mu}\right)$ for the material of the given prism.

We have $\mathrm{r}+\mathrm{r}^{\prime}=\mathrm{A}$

$\therefore \quad r=\left(A-\sin ^{-1} \dfrac{1}{\mu}\right)$

Also, $\mathrm{i}+\mathrm{e}\left(+\mathrm{i}+90^{\circ}\right)=\mathrm{A}+\mathrm{D}$

$\therefore \quad \mathrm{i}=(\mathrm{A}+\mathrm{D})-90^{\circ}$

$\therefore \quad \cos i=\cos \left[(A+D)-90^{\circ}\right]=\cos \left[90^{\circ}-(A+D)\right]$

$=\sin (\mathrm{A}+\mathrm{D})$

$\therefore \quad \mathrm{i}=\cos ^{-1}[\sin (\mathrm{A}+\mathrm{D})]$

Again, by Snell’s law,

$\dfrac{\sin i}{\sin r}=\mu$

$\therefore \sin \mathrm{r}=\dfrac{\sin \mathrm{i}}{\mu}$

$=\sin \left[\cos ^{-1}(\sin (A+D))\right]$

$\therefore \mathrm{r}=\sin ^{-1}\left[\sin \left\{\cos ^{-1}(\sin (\mathrm{A}+\mathrm{D}))\right\}\right]$

Difficult

Refraction of Light by Prism

62. Light, incident on the face $A B$, of an equilateral prism (made from a material of redfractive index $\mu$ ), undergoes total internal reflection at the base $\mathrm{BC}$ and then falls on the other face $A C$ of the prism. The angle of emergence, from the face $\mathrm{AC}$, would equal

(1) $\sin ^{-1}\left[\mu \sin \left(\dfrac{\pi}{3}+\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

(2) $\sin ^{-1}\left[\mu \sin \left(\dfrac{\pi}{3}-\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

(3) $\sin ^{-1}\left[\sin \left(\dfrac{\pi}{3}+\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

(4) $\sin ^{-1}\left[\sin \left(\dfrac{\pi}{3}-\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

Show Answer

Correct answer: (2)

Solution:

Since there is total internal reflection at the face $\mathrm{BC}$, we have

$\sin \theta=\dfrac{1}{\mu} \text { or } \theta=\sin ^{-1}\left(\dfrac{1}{\mu}\right)$

From cyclic quadrilaterial $\mathrm{QRCS}$, we get

$<\mathrm{ROS}=180^{\circ}-<\mathrm{C}=120^{\circ}$

$\therefore$ From triangle ROS, we get

$<\mathrm{r} _{2}=180^{\circ}-\left[120^{0}+\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right]=\left(60^{0}-\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)$

But $\dfrac{\sin \mathrm{e}}{\sin \mathrm{r} _{2}}=\mu$

$\therefore e=\sin ^{-1}\left[\mu \sin \left(60^{0}-\sin ^{-1}\left(\dfrac{1}{\mu}\right)\right)\right]$

Difficult

Real and Apparent Height

63. A transparent oil, of redfractive index $7 / 5$, is poured over a layer of water (redfractive index $=4 / 3$ ) in a glass container. The ‘depths’ of the oil and water layers, are in the inverse ratio of their redfractive indices.

A tiny shining metal ball, lying at the bottom of the container, is viewed from above. Its ‘apparent height’, above the bottom of the container equals, a dfraction, $f$, of the total depth of the two liquids in the container. The dfraction, $f$, is (nearly) equal to

(1) 0.27

(2) 0.30

(3) 0.32

(4) 0.35

Show Answer

Correct answer: (1)

Solution:

The depths, of the two liquids, are in the ratio $\dfrac{4}{3}: \dfrac{7}{5} ;$ i.e., $20: 21$. We can therefore, take the total depth, of the two liquids, as $[(20+21) x]$ units, i.e.; (41x) units.

Apparent depth of the top layer $=\left(20 x \div \dfrac{7}{5}\right)=\dfrac{100 x}{7}$ units

Apparent depth of the bottom layer $=\left(21 x \div \dfrac{4}{3}\right)=\dfrac{63 x}{4}$ units

$\therefore$ Total apparent depth $=\left(\dfrac{841}{28} x\right)$ units

$\therefore$ Apparent height of the ball $=\left(41 x-\dfrac{841}{28} x\right)$ units

$=\dfrac{307}{28} x$ units

$\therefore \mathrm{f}=\dfrac{307}{28} x\left(\dfrac{1}{41 x}\right)=\left(\dfrac{307}{1148}\right) \simeq 0.27$

Difficult

Refractive Index

64. A tiny air bubble, trapped inside a glass cube of side a, appears to be at distances $x _{1}$ and $x _{2}$, respectively, when viewed from two opposite faces of the cube. The redfractive index, of glass, and the actual distance $(x)$ of the air bubble, from the first of the two faces, are then equal, respectively, to

(1) $\left(\dfrac{x _{1}+x _{2}}{\mathrm{a}}\right)$ and $\left(\dfrac{\mathrm{a} x _{1}}{x _{1}+x _{2}}\right)$

(2) $\left(\dfrac{x _{1}+x _{2}}{\mathrm{a}}\right)$ and $\left(\dfrac{\mathrm{a} x _{2}}{x _{1}+x _{2}}\right)$

(3) $\left(\dfrac{\mathrm{a}}{x _{1}+x _{2}}\right)$ and $\left(\dfrac{\mathrm{a} x _{2}}{x _{1}+x _{2}}\right)$

(4) $\left(\dfrac{\mathrm{a}}{x _{1}+x _{2}}\right)$ and $\left(\dfrac{\mathrm{a} x _{1}}{x _{1}+x _{2}}\right)$

Show Answer

Correct answer: (4)

Solution:

The apparent distance, of the air bubble from the first face, is $\dfrac{1}{\mu}$ times its real distance $(=x)$ from that face.

Hence $x _{1}=\dfrac{x}{\mu}$

For the opposite face, the real distance, of the air bubble, would be (a-x). Hence $x _{2}=\left(\dfrac{\mathrm{a}-x}{\mu}\right)$

Dividing, we get

$\begin{aligned} & \dfrac{x _{1}}{x _{2}}=\dfrac{x}{\mathrm{a}-x} \\ \therefore & \dfrac{x _{1}}{x _{1}+x _{2}}=\dfrac{x}{x+(\mathrm{a}-x)}=\dfrac{x}{\mathrm{a}} \\ \therefore \quad & x=\mathrm{a}\left(\dfrac{x _{1}}{x _{1}+x _{2}}\right) \\ \therefore \quad & \mu=\dfrac{x}{x _{1}}=\left(\dfrac{\mathrm{a} x _{1}}{x _{1}+x _{2}}\right) \dfrac{1}{x _{1}} \\ & =\dfrac{\mathrm{a}}{\left(x _{1}+x _{2}\right)} \end{aligned}$

Average

‘Real and Apparent Depth’

65. Three layers of three different transparent media (of redfractive indices $\mathbf{n} _{1}, \mathbf{n} _{2}$ and $\mathbf{n} _{3}$, respectively) have thicknesses $d _{1}, d _{2}$ and $d _{3}$, respectively. Their total apparent depth, when viewed from above, would appear to be (nearly)

(1) $\dfrac{\mathrm{d} _{1}}{\mathrm{n} _{2}}+\dfrac{\mathrm{d} _{2}}{\mathrm{n} _{3}}+\dfrac{\mathrm{d} _{3}}{\mathrm{n} _{1}}$

(2) $\mathrm{d} _{1}\left(1-\dfrac{1}{\mathrm{n} _{1}}\right)+\mathrm{d} _{2}\left(1-\dfrac{1}{\mathrm{n} _{2}}\right)+\mathrm{d} _{3}\left(1-\dfrac{1}{\mathrm{n} _{3}}\right)$

(3) $\dfrac{1}{\mathrm{n} _{1} \mathrm{n} _{2} \mathrm{n} _{3}}\left(d _{1} \mathrm{n} _{2} \mathrm{n} _{3}+d _{2} \mathrm{n} _{3} \mathrm{n} _{1}+d _{3} \mathrm{n} _{1} \mathrm{n} _{2}\right)$

(4) $\dfrac{1}{\mathrm{n} _{1} \mathrm{n} _{2} \mathrm{n} _{3}}\left(d _{1} \mathrm{n} _{1} \mathrm{n} _{2}+d _{2} \mathrm{n} _{2} \mathrm{n} _{3}+d _{3} \mathrm{n} _{3} \mathrm{n} _{1}\right)$

Show Answer

Correct answer: (3)

Solution:

The apparent depth, of a layer of a medium of redfractive index $\mu$, is $\left(\dfrac{1}{\mu}\right)$ times its real depth. Hence the total apparent depth, of the three layers, would be

$\begin{aligned} \mathrm{d}^{\prime}= & \left.\dfrac{\mathrm{d} _{1}}{\mathrm{n} _{1}}+\dfrac{\mathrm{d} _{2}}{\mathrm{n} _{2}}+\dfrac{\mathrm{d} _{3}}{\mathrm{n} _{3}}\right) \\ & =\dfrac{1}{\mathrm{n} _{1} \mathrm{n} _{2} \mathrm{n} _{3}}\left(\mathrm{~d} _{1} \mathrm{n} _{2} \mathrm{n} _{3}+\mathrm{d} _{2} \mathrm{n} _{3} \mathrm{n} _{1}+\mathrm{d} _{3} \mathrm{n} _{1} \mathrm{n} _{2}\right) \end{aligned}$

Easy

P–N Junction Diode

66. The reverse saturation current in a $p-n$ junction diode

(1) increases with increase in applied voltage

(2) decreases with increase in applied voltage

(3) increases with increase in temperature

(4) decreases with increase in temperature

Show Answer

Correct answer: (3)

Solution:

Reverse saturation current depends on the rate of generation of electron hole pair, which is independent of the applied voltage.

In fact, $I _{S} \alpha \exp \left(\dfrac{-E _{g}}{k T}\right)$

$\left(\mathrm{E} _{\mathrm{g}}=\right.$ energy band gap, $\mathrm{K}=$ Boltzman constant, $\mathrm{T}=$ Temperature)

Hence $I _{S}$ increases as $T$ increases.

Average

P–N Junction Diode

67. The slope of $(\log I)$ vs voltage curve (for a $p-n$ junction diode), can be used to determine,

(1) Planck’s constant

(2) Boltzman constant (Correct)

(3) Reverse saturation current (Incorrect)

(4) Mobility of minority charge carriers

Show Answer

Correct answer: (2)

Solution:

We have

$\begin{aligned} \mathrm{I}=\mathrm{I} _{0} & \left\{\exp \left(\dfrac{\mathrm{eV}}{\eta \mathrm{kT}}\right)\right\} \\ & \Rightarrow \log \mathrm{I}=\log \mathrm{I} _{\mathrm{S}}+\dfrac{\mathrm{eV}}{\eta \mathrm{kT}} \\ & \Rightarrow \text { slope }=\dfrac{\mathrm{e}}{\eta \mathrm{kT}} \end{aligned}$

It can give value of $k$.

Easy

P–N Junction Diode

68. The depletion region in a $\mathbf{p}-\mathbf{n}$ junction is so called because it is deplected of,

(1) Ionized donor impurities

(2) Ionized acceptor impurities

(3) Immobile charge carriers

(4) Mobile charges carriers

Show Answer

Correct answer: (4)

Solution:

There is a recombination of electrons and holes in the deplections region. Hence the charge carriers, in this region, are immobile.

Easy

$P-N$ Junction Diode

69. The motion of charge carriers, across the potential barrier, in a forward and reverse biased $\mathbf{p}-\mathbf{n}$ junction, is due, respectively, to

(1) Drift, diffusion

(2) Diffusion, drift

(3) Drift in both cases

(4) Diffusion in both cases

Show Answer

Correct answer: (2)

Solution:

The charge carriers, diffuse across the junction during the forward biasing of a $\mathbf{p}-\mathbf{n}$ junction.

During reverse biasing, the motion of charge cariers, is due to their drift.

Average

Zener Diode

70. In the circuit given below, the Zener voltage is maintained at $6 \mathrm{~V}$. Zener resistance is negaligible and the Zener knee current is $10 \mathrm{~mA}$. The minimum value of $R _{L}$, is

(1) $75 \Omega$

(2) $150 \Omega$

(3) $300 \Omega$

(4) $600 \Omega$

Show Answer

Correct answer: (4)

Solution:

We have,

$\mathrm{I} _{\mathrm{S}}=\dfrac{9 \mathrm{~V}-6 \mathrm{~V}}{150 \Omega}=\dfrac{3}{150} \mathrm{~A}=20 \mathrm{~mA}$

$\mathrm{I} _{\mathrm{S}}=\mathrm{I} _{\mathrm{Z}}+\mathrm{I} _{\mathrm{L}}$

$\Rightarrow 20 \mathrm{~mA}=10 \mathrm{~mA}+\mathrm{I} _{\mathrm{L}}$

$\Rightarrow \mathrm{I} _{\mathrm{L}}=10 \mathrm{~mA}$

$\Rightarrow \mathrm{R} _{\mathrm{L}}=\dfrac{\mathrm{V} _{\mathrm{Z}}}{10 \mathrm{~mA}}=\dfrac{6 \mathrm{~V}}{10 \mathrm{~mA}}=600 \Omega$

Average

Zener Diode

71. In the circuit given below, the Zener voltage is $6 \mathrm{~V}$, dynamic resistance of Zener is $10 \Omega$ and Zener current is negligible. If input voltage varies between $9 \mathrm{~V}$ to $12 \mathrm{~V}$, the output voltage would

(1) always be $6 \mathrm{~V}$

(2) lie between $9 \mathrm{~V}$ to $12 \mathrm{~V}$

(3) equal $9 \mathrm{~V}$

(4) lie between $6.1875 \mathrm{~V}$ to $6.375 \mathrm{~V}$

Show Answer

Correct answer: (4)

Solution:

The equivalent circuit diagram gives

$\begin{aligned} & I=\dfrac{V _{i}-V _{z}}{150+10}=\dfrac{V _{i}-6}{160} \\ & V _{0}=V _{z}+10 I \\ & =6 V _{z}+10\left(\dfrac{V _{i}-6}{160}\right) \end{aligned}$

If $\mathrm{V} _{1}=9 \mathrm{~V} \Rightarrow \mathrm{V} _{0}=6.1875 \mathrm{~V}$

If $\mathrm{V} _{1}=12 \mathrm{~V} \Rightarrow \mathrm{V} _{0}=6.375 \mathrm{v}$

Easy

Doping of a Diode

72. It will be correct to say that,

(1) A lightly doped diode has a thin deplection layer and a higher breakdown voltage (Incorrect)

(2) A heavily doped diode has a thin deplection layer and a lower breakdown voltage (Correct)

(3) A lightly doped diode has a wide deplection layer and a lower breakdown voltage (Incorrect)

(4) A heavily doped diode has wide deplection layer and a higher breakdown voltage Incorrect)

Show Answer

Correct answer: (2)

Solution:

The width of the deplection layer, as well as the barrier potential across the junction, both decrease with an increase in the extent of doping, of the $\mathrm{p}$ and $\mathrm{n}$ sides, of the $\mathrm{p}-\mathrm{n}$ junction diode.

Easy

Junction Transistor

73. The input resistance, of a pnp transistor, in its common emitter configuration, is given by,

(1) $\left(\dfrac{\Delta \mathrm{V} _{\mathrm{BE}}}{\Delta \mathrm{I} _{\mathrm{B}}}\right)$ at constant $\mathrm{V} _{\mathrm{CE}}$

(2) $\left(\dfrac{\Delta \mathrm{V} _{\mathrm{BE}}}{\Delta \mathrm{I} _{\mathrm{B}}}\right)$ at constant $\mathrm{I} _{\mathrm{C}}$

(3) $\left(\dfrac{\Delta \mathrm{V} _{\mathrm{CE}}}{\Delta \mathrm{I} _{\mathrm{C}}}\right)$ at constant $\mathrm{V} _{\mathrm{B}}$

(4) $\left(\dfrac{\Delta \mathrm{V} _{\mathrm{CE}}}{\Delta \mathrm{I} _{\mathrm{C}}}\right)$ at constant $\mathrm{V} _{\mathrm{BE}}$

Show Answer

Correct answer: (1)

Solution:

The input resitance is defined as

$\mathrm{r} _{\mathrm{i}}=\left(\dfrac{\Delta \mathrm{V} _{\mathrm{BE}}}{\Delta \mathrm{I} _{\mathrm{B}}}\right) \text { at constant } \mathrm{V} _{\mathrm{CE}}$

Easy

Junction Transistor

74. In a bijunction transistor, the base region is,

(1) Lightly doped so that recombination is more

(2) Lightly doped so that recombination is less

(3) Heavily doped so that recombination is more

(4) Heavily doped so that recombination is less

Show Answer

Correct answer: (2)

Solution:

The base region is kept thin and is also lightly doped. This is to ensure that the charge carriers, from the emitter, cross over (without any significant recombination) and get collected in the collector region.

Average

Junction Transistor

75. One of the differences/similarity, in the operation of a transistor amplifier, and a step-up transformer, is

(1) In transistor, if voltage increases, current also increases, in step-up transformer, if voltage increases, current decreases

(2) In transistor, if voltage increases, current increase; in step-up transformer, if voltage increasee, current increases

(3) In both of them, if voltage increases, current also increases

(4) In both of them, if voltage increases, current decreases

Show Answer

Correct answer: (1)

Solution:

The transistor can bring about an increases in voltage along with an increase in current. This is because the increased power, in the ‘output’ of a transistor, comes at the expense of the power supplied by the biasing battery.

In a step-up transformer, there is no such power source. Hence the ‘output’ power - even in the case of an ideal transformer - can at best be equal to the ‘input’ power. The increase in voltage, here, is accompnied by a decrease in current, and vice-versa.

Average

‘Current Gain’ in a Transistor

76. For a given transistor, if $\beta=200$, the value of $\alpha$ will be (almost) equal to

(1) 0.0

(2) 1.0

(3) 200

(4) 201

Show Answer

Correct answer: (2)

Solution:

$\alpha=\dfrac{\beta}{1+\beta}=\dfrac{200}{201}=0.995 \simeq 1.0$

This is so because $\alpha$ is the current gain in the common base configuration. It equals $\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{E}}}$ and is always less than unity.

’ $\beta$ ‘, on the other hand, is the current gain in the common-emitter configuration. It equals $\dfrac{\Delta \mathrm{I} _{\mathrm{C}}}{\Delta \mathrm{I} _{\mathrm{B}}}$. Remembering that $\Delta \mathrm{I} _{\mathrm{E}}=\left(\Delta \mathrm{I} _{\mathrm{C}}+\Delta \mathrm{I} _{\mathrm{B}}\right)$, we get the given relation between $\alpha$ and $\beta$.

Average

Multimeter Use

77. While checking a diode with a digital multimeter, set to its ‘diode’ symbol, a good diode is expected to show.

(1) $0 \mathrm{~V}$ in both forward and reverse biasing

(2) $0.5 \mathrm{~V}-0.7 \mathrm{~V}$ in both forward and reverse biasing

(3) $0.5 \mathrm{~V}-0.7 \mathrm{~V}$ in only one direction of polarity

(4) 1 in both forward and reverse polarity

Show Answer

Correct answer: (3)

Solution:

In forward bias, it shows voltage drop across the diode which is small, $0.5-0.7 \mathrm{~V}$. In reverse bias, it shows large, overload, signifying large $(\infty)$ resistance.

Easy

LED

78. The voltage drop, across a LED, when connected in forward bias configuration to a multimeter, is,

(1) 0.5 to $0.7 \mathrm{mV}$

(2) 1.5 to $2.2 \mathrm{~V}$

(3) 0.5 to $0.7 \mathrm{~V}$

(4) 1.5 to $2.2 \mathrm{mV}$

Show Answer

Correct answer: (2)

Solution:

The voltage drop, across LED, in the forward bias configuration, is in the range of $1.5 \mathrm{~V}$ to $2.2 \mathrm{~V}$.

Easy

Identification of a Given Device

79. A small device has a yellow colored capsule top and has 220 written on it. There is no + or - sign on it. This device is likely to be

(1) Electrolyte capacitor of value $220 \mu \mathrm{F}$

(2) Electrolytc capacitor of value $22 \rho \mathrm{F}$

(3) Ceramic capacitor of value $220 \mu \mathrm{F}$

(4) Ceramic capacitor of value $22 \mathrm{pF}$

Show Answer

Correct answer: (4)

Solution:

The given description matches a ceramic capacitor of value $22 \mathrm{pF}$.

Average

Identifying the Base of a Transistor

80. If negative probe of a multimeter is placed $n$ the base of a $\mathrm{p}-\mathrm{n}-\mathrm{p}$ transistor, then ohm meter reading, with other probe placed on one of the other pins, will show

(1) a low resistance

(2) a high resistance

(3) zero resistance

(4) (1-) reading

Show Answer

Correct answer: (1)

Solution:

When the negative probe of a multimeter (adjusted to work as an ohm meter) is placed on the base of a $\mathrm{p}-\mathrm{n}-\mathrm{p}$ transistor, and the other probe is placed on the of the other pins, it will show a ’low resistnace’ reading. This is so because this way, of connecting the multimeter probes, corresponds to a forward biasing of the ( $\mathrm{p}-\mathrm{n}$ junction) being investigated.