n divisions of vernier scale =(n-1) divisions of main scale.

eg; 10 divisions on vernier scale(V) = 9 divisions on main sale.

1 division of $V=\frac{9}{10}$ division of $S$.

s=1 mm, V=0.9 mm.

The bottom scale is the main scale.

The top scale is a movable scale, is called the Vernier scale.

The 0 of the Vernier scale is not matching with the 0 of the main scale.

AB = AC - BC

AB = 6S - 6V

Least count of Vernier Calliper =S-V

$n V=(n-1) s$

$V=\frac{n-1}{n} s$

Least count of Vernier Calliper $=S-\frac{n-1}{n} S=\frac{S}{n}$.

$s=1 \mathrm{mm}$

0 of Vernier occurs after 2 readings of main scale.

Main scale reading = 2 mm

The 4th reading of V coincides with main scale.

$x=4S-4 V=4 \times 1-4 \times 0.9= 0.4 mm$

The total reading of that length which the Vernier caliper is showing to us,

2 + 0.4 = 2.4 mm

In a Vernier caliper there is a moving jaw which is adjusted.

You find how many divisions of the main scale are and then, matching with how many divisions on the Vernier scale.

The Vernier scale is at the bottom.

The number of divisions on the Vernier scale are 50 and they match with 49 divisions of the main scale.

The main scale reading is 3.3 centimeters.

The Vernier scale we have the 20th reading.

2 Vernier Callipers in which on mainscale 1 cm $\rightarrow$ 10 divisions

$V_1: S_1=1 \mathrm{~mm}$

$V_2: S_2=1 \mathrm{~mm}$

Second Calliper :10 divisions of V = 11 divisions of S

V = 1.1 mm

$C_1$: Main Scale Reading $=2.8 \mathrm{cm}$

$x$ has to be added to 2.8

$x =7 \mathrm{S}-7 \mathrm{V}$

$=7 \times 1 \mathrm{mm}-7 \times 0.9 \mathrm{mm}$

$=7- 6.3$

$=0.7 \mathrm{mm}$

The total reading by C1 $=2.8 \mathrm{cm}+0.7 \mathrm{mm}$

$=2.87 \mathrm{cm}$

Second case: $x_1$ has to be measured.

$x_1=$ distance between main scale reading and zero of vernier.

10 readings of the Vernier were matching 11 readings of the main scale.

$x_1$ $=8 \mathrm{S}-7 \mathrm{V}$

$=8 \mathrm{mm}-7 \times 1.1 \mathrm{mm}$

$=(8-7.7) \mathrm{mm}$

$=0.3 \mathrm{~mm}$

$C_2$ $\text { reading }=2.8 \mathrm{cm}+0.3 \mathrm{mm}=2.83 \mathrm{cm}$

Thread of a screw has a helical shape which implies, if we make rotation on the thread of screw, we also make an axial movement.

The axial distance moved when we complete 1 revolution on screw = pitch

Pitch - Linear dimension corresponds to $360^{\circ}$ rotation on a screw.

The linear scale can measure the axial distance which the screw moves.

The partial rotations of the screw are measured by circular scale.

When these two are added we get the total reading.

Pitch of the screw corresponds to the smallest reading of main scale. We rotate the spindle by 1 revolution and observe the axial distance moved on the mainscale.

Typically ~ 0.5 mm or 1 mm

Circular scale enables us to read fractional reading.

Let number of divisions on circular scale be n.

Let pitch of the screw be p.

Least Count of the screw = $\frac{p}{n}$

If Circular scale reading = m.

Extra distance / length to be added to main scale reading = ${m} \times\frac{p}{n}$.

For example, pitch p = 0.5 mm

Number of divisions on the circular scales n = 50

Least Count = $\frac{0.5}{50}$ = 0.01 mm

The pitch of this instrument is 0.5 millimeters.

The circular scale has 50 divisions.

The least count of this instrument is 0.01 mm.

Zero error has to be substracted from the reading.

Zero error can be the positive (+ve) or negative (-ve).

ACCURACY :- how close we are to the actual numerical value.

PRECISION :- resolution of the quantity measured.

If we take measurements from a typical scale $\rightarrow$ precision 1 mm.

Accuracy depends on many factors include precision.

Eg; True Length of a line = 3.678 cm.

Measure with scale of Least Count 1 mm = 3.5 cm.

Measure with an instrument of Least Count 0.1 mm = 3.38 cm.

Errors tend to be in 1 direction, i.e., they are either the positive or negative.

a) Instrumental Error :- Calibration of instrument is not perfect, zero error.

Occur irregularly, are random unpredictable fluctuations in instruments, (temperature, voltage, vibrations in set up)

The smallest resolution of the instrument.

The least count of a vernier caliper is 0.1 mm

Repeat observations.

Take the arithmetic mean of these readings.

We take arithmetic mean as the actual value.

Absolute Error

Relative Error

Percentage Error

n measurements

$a_1, a_2, \ldots a_n$

$a_{\text {mean }}=\frac{a_1 + a_2 \ldots+a_n}{n}=\frac{\sum_{i=1}^n a_i}{n}$

Magnitude of difference between individual measurement and true measurement = absolute error

$\left|\Delta a_i\right|=\left|a_i-a_{\text {mean }}\right|$

$a_1 \ldots a_n$

true value not known, mean is taken as true value.

$a_{\text {mean }}=\frac{a_1+a_2 \ldots+a_n}{n}$

$\Delta a_1=a_1-a_{\text {mean }}$

$\Delta a_2=a_2-a_{\text {mean }}$

$\therefore a_n=a_n-a_{\text {mean }}$

Take absolute values.

$|\Delta a| \leftarrow$ always positive.

Arithmetic mean of all absolute errors, called mean absolute error of $a$.

$\Delta a_{\text {mean }}=\frac{\left(\left|\Delta a_1\right|+\left|\Delta a_2\right| \cdots+|| a_n \mid\right)}{n}$

Single measurement, we expect this to be in range.

$a_{\text {mean }} \pm \Delta a_{\text {mean }}$

$a_{\text {mean }}-\Delta a_{\text {mean }} \leqslant a \leqslant a_{\text {mean }} +\Delta a_{\text {mean }}$

Relative error $=\frac{\Delta a_{\text {mean }}}{a_{\text {mean }}}$

Relative error is always a fraction.f

% error : $\delta a=\frac{\Delta a_{\text {mean }}}{a_{\text {mean }}} \times 100$%

Example: Time period of oscillations

5 measurements

$2.63 \mathrm{s}$, $2.56 \mathrm{s}$, $2.42 \mathrm{s}$, $2.71 \mathrm{s}$, and $2.80 \mathrm{s}$

Total sum = $\sum 13.12 \mathrm{s}$

$\text { mean } =\frac{13.12}{5}$

$=2.624 \mathrm{s}=2.62 \mathrm{s}$

$\Delta a_1=2.63-2.62=0.01 \mathrm{s}$

$\Delta a_2=2.56-2.62=-0.06 \mathrm{s}$

$\Delta a_3=2.43-2.62=-0.20 \mathrm{s}$

$\Delta a_4=2.71-2.62=0.09 \mathrm{s}$

$\Delta a_3=2.80-2.62=0.18 \mathrm{s}$

absolute error $=\frac{(0.01+0.06+0.20+0.09+0.18)}{ 5}$

Absolute error = $\frac{0.54}{5}=0.11 \mathrm{~s}$

$T=2.62 \pm 0.11 \mathrm{~s}$

$\text { error }>0.1 \mathrm{~s}$

$T=2.6 \pm 0.1 \mathrm{~s}$

relative error % $= \frac{0.1}{2.6} \times 100=4$%