Thinking Process

When the pebble is dropped from the top, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant.

Answer (c) When the pebble is falling through the viscous oil the viscous force is

$$ F=6 \pi \eta r v $$

where $r$ is radius of the pebble, $v$ is instantaneous speed, $\eta$ is coefficient of viscosity. As the force is variable, hence acceleration is also variable so $v$ - $t$ graph will not be straight line.First velocity increases and then becomes constant known as terminal velocity.

Answer (a) Consider the diagram where an ideal fluid is flowing through a pipe.

As given

$d_1=$ Diameter at 1 st point is 2.5 .

$d_2=$ Diameter at 2nd point is 3.75 .

Applying equation of continuity for cross-sections $A_1$ and $A_2$.

$$ \begin{aligned} \Rightarrow \quad A_1 v_1 & =A_2 v_2 \\ \Rightarrow \quad \frac{v_1}{v_2} & =\frac{A_2}{A_1}=\frac{\pi(r_2^{2})}{\pi(r_1^{2})}=(\frac{r_2}{r_1})^{2} \\ & =(\frac{\frac{3.75}{2}}{\frac{2.5}{2}})^{2}=(\frac{3.75}{2.5})^{2}=\frac{9}{4}[\begin{matrix} r_2=\frac{d_2}{2} \\ r_1=\frac{d_1}{2} \end{matrix} ] \end{aligned} $$

Answer (c) According to the question, the observed meniscus is of convex figure shape. Which is only possible when angle of contact is obtuse. Hence, the combination will be of mercury-glass $(140^{\circ})$

Answer $(b, d)$

Consider the diagram where two molecules of a liquid are shown. One is well inside the liquid and other is on the surface. The molecule $(A)$ which is well inside experiences equal forces from all directions, hence net force on it will be zero.

And molecules on the liquid surface have some extra energy as it surrounded surraind by only lower half side of liquid molecules.

Answer $(b, c)$

Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

As magnitude of component is considered, hence, it will not have any direction. So, pressure is a scalar quantity.

Thinking Process

When any body floats in a liquid, the upthrust force acting on the body due to the displaced liquid is balanced by its weight.

Answer $(a, b)$

When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, hence upthrust force will also decrease which is only possible when $l$ decreases.

As $l$ decreases volume of water displaced by the block decreases, hence $h$ decreases.

Note As the coin falls into water, it displaces some volume of water which is very less hence, we neglect volume of the coin.

Answer (c, $d)$

For liquids coefficient of viscosity, $\quad \eta \propto \frac{1}{\sqrt{T}}$

i.e., with increase in temperature $\eta$ decreases.

For gases coefficient of viscosity, $\quad \eta \propto \sqrt{T}$

i.e., with increase in temperature $\eta$ increases.

Answer (b, c)

Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be velocity gradient hence each line of flow can be easily differentiated. Also higher the coefficient of viscosity lower will be Reynolds number, hence flow more like to be streamline.

Answer No, surface tension is a scalar quantity.

Surface tension $=\frac{\text { Work done }}{\text { Surface area }}$, where work done and surface area both are scalar quantities.

Answer Given, density of ice $(\rho_{\text {ice }})=0.917 g / cm^{3}$

Density of water $(\rho_{w})=1 g / cm^{3}$

Let $V$ be the total volume of the iceberg and $V^{\prime}$ of its volume be submerged in water. In floating condition.

Weight of the iceberg $=$ Weight of the water displaced by the submerged part by ice

or

$$ \begin{aligned} V \rho_{ice} g & =V^{\prime} \rho_{w} g \\ \frac{V^{\prime}}{V} & =\frac{\rho_{ce}}{\rho_{w}}=\frac{0.917}{1}=0.917 \quad(\because \text { Weight }=m g=v \rho g) \end{aligned} $$

Answer Consider the diagram,

The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed in water, then the reading of the scale will be equal to the thrust on the block due to water.

Thrust $=$ weight of water displaced

$=V \rho_{w} g$ (where $V$ is volume of the block and $\rho_{w}$ is density of water)

$$ =\frac{m}{\rho} \rho_{w} g=(\frac{\rho_{w}}{\rho}) m g $$

$(\because.$ Density of the block $.\rho=\frac{\text { mass }}{\text { volume }}=\frac{m}{V})$

Thinking Process

As the elevator is accelerating upward the net acceleration of the block with respect to the elevator can be calculate by the concept of pseudo force.

Answer Consider the diagram.

Let the density of water be $\rho_{w}$ and a cubical block of ice of side $L$ be floating in water with $x$ of its height $(L)$ submerged in water.

$$ \begin{aligned} \text { Volume of the block }(V) & =L^{3} \\ \text { Mass of the block }(m) & =V \rho=L^{3} \rho \\ \text { Weight of the block } & =m g=L^{3} \rho g \end{aligned} $$

1st case

Volume of the water displaced by the submerged part of the

$\therefore$ Weight of the water displaced by the block

In floating condition, $x L^{2} \rho_{w} g$

Weight of the block $=$ Weight of the water displaced by the block

$$ \begin{matrix} L^{3} \rho g & =x L^{2} \rho_{w} g \\ \text { or } & \frac{x}{L}=\frac{\rho}{\rho_{w}}=x \end{matrix} $$

2nd case

When elevator is accelerating upward with an acceleration a, then effective acceleration

Then, weight of the block

$$ \begin{aligned} & =(g+a) \\ & =m(g+a) \\ & =L^{3} \rho(g+a) \\ & L^{3} \rho(g+a)=(x_1 L^{2}) \rho_{w}(g+a) \\ & \frac{x_1}{L}=\frac{\rho}{\rho_{w}}=x \end{aligned} $$

( $\because$ Pseudo force is downward)

Let $x_1$ fraction be submerged in water when elevator is accelerating upwards.

Now, in the floating condition, weight of the block = weight of the displaced water

or

From 1st and 2nd case,

We see that,the fraction of the block submerged in water is independent of the acceleration of the elevator.

Note We should not confuse with the concept of pseudo force, i.e., pseudo force is downward, hence fraction will change due to increased force.

Answer Given, radius $(r)=2.5 \times 10^{-5} m$

Surface tension $(S)=7.28 \times 10^{-2} N / m$

Angle of $contact(\theta)=0^{\circ}$

The maximum height to which sap can rise in trees through capillarity action is given by

$$ \begin{aligned} h & =\frac{2 S \cos \theta}{r \rho g} \text { where } S=\text { Surface tension, } \rho=\text { Density, } r=\text { Radius } \\ & =\frac{2 \times 7.28 \times 10^{-2} \times \cos 0^{\circ}}{2.5 \times 10^{-5} \times 1 \times 10^{-3} \times 9.8}=0.6 m \end{aligned} $$

This is the maximum height to which the sap can rise due to surface tension. Since, many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.

Thinking Process

As the tanker starts accelerating, free surface of the tanker will not be horizontal because pseudo force acts.

Answer Consider the diagram where a tanker is accelerating with acceleration a.

Consider an elementary particle of the fluid of mass $d m$.

The acting forces on the particle with respect to the tanker are shown above .

Now, balancing forces (as the particle is in equilibrium) along the inclined direction component of weight $=$ component of pseudo force $d m g \sin \theta=d m a \cos \theta$ (we have assumed that the surface is inclined at an angle $\theta$ ) where, $d m$ is pseudo force

$$ \begin{matrix} \Rightarrow & g \sin \theta & =a \cos \theta \\ \Rightarrow & & a & =g \tan \theta \\ \Rightarrow & & \tan \theta & =\frac{a}{g}=\text { slope } \end{matrix} $$

Thinking Process

In this process, conservation of mass will be applied. Before and after collapse, mass and hence, volume of the system remains conserved.

Answer Consider the diagram.

Radii of mercury droplets

$$ \begin{aligned} & r_1=0.1 cm=1 \times 10^{-3} m \\ & r_2=0.2 cm=2 \times 10^{-3} m \end{aligned} $$

Surface tension $(T)=435.5 \times 10^{-3} N / m$

Let the radius of the big drop formed by collapsing be $R$.

$\therefore$ Volume of big drop $=$ Volume of small droplets

or

$$ \begin{aligned} \frac{4}{3} \pi R^{3} & =\frac{4}{3} \pi r_1^{3}+\frac{4}{3} \pi r_2^{2} \\ R^{3} & =r_1^{3}+r_2^{3} \\ & =(0.1)^{3}+(0.2)^{3} \\ & =0.001+0.008 \\ & =0.009 \end{aligned} $$

$$ \quad R^{3}=r_1^{3}+r_2^{3} $$

or

$$ R=0.21 cm=2.1 \times 10^{-3} m $$

$\therefore$ Change in surface area

$$ \Delta A=4 \pi R^{2}-(4 \pi r_1^{2}+4 \pi r_2^{2}) $$

$$ =4 \pi[R^{2}-(r_1^{2}+r_2^{2})] $$

$$ \begin{aligned} \therefore \quad \text { Energy released }= & T \cdot \Delta A \quad(\text { where } T \text { is surface tension of mercury) } \\ & =T \times 4 \pi[R^{2}-(r_1^{2}+r_2^{2})] \\ & =435.5 \times 10^{-3} \times 4 \times 3.14[(2.1 \times 10^{-3})^{2}. \\ & ..=435.5 \times 4 \times 3.14[4.41-5] \times 10^{-6} \times 10^{-3}+4 \times 10^{-6})] \\ & =-32.23 \times 10^{-7} \quad \text { (Negative sign shows absorption) } \end{aligned} $$

Therefore, $3.22 \times 10^{-6} J$ energy will be absorbed.

Note In this process, energy is not conserved. Energy is Ist due to the collapsing, in the form of radiations.

Answer When a big drop of radius $R$, breaks into $N$ droplets each of radius $r$, the volume remains constant.

$$ \therefore \quad \quad\text{Volume of big drop} =N \times \text{volume of each small drop}\\ \frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3\\ \text{or} \quad \quad R^3 =Nr^3\\ \text{or} \quad \quad N=\frac{R^3}{r^3}\\ \text{Now}, \quad \quad \text{change in surface area} = 4 \pi R^2 - N4 \pi r^2\\ =4 \pi (R^2-nr^2)\\ \text{Energy released} = T \times \Delta A = S \times 4 \pi (R^2 -Nr^2) \quad \quad \quad [T= \text{Surface tension}]$$

Due to releasing of this energy, the temperature is lowered.

If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\Delta \theta$, then energy released $=m s \Delta \theta$ [s $=$ specific heat $\Delta \theta=$ change in temperature]

$$ \begin{aligned} T \times 4 \pi(R^{2}-N r^{2}) & =(\frac{4}{3} \times R^{3} \times \rho) s \Delta \theta \\ \Delta \theta & =\frac{T \times 4 \pi(R^{2}-N r^{2})}{\frac{4}{3} \pi R^{3} \rho \times s} \\ & =\frac{3 T}{\rho s}[\frac{R^{2}}{R^{3}}-\frac{N r^{2}}{R^{3}}] \\ & =\frac{3 T}{\rho s}[\frac{1}{R}-\frac{(R^{3} / r^{3}) \times r^{2}}{R^{3}}] \\ & =\frac{3 T}{\rho s}[\frac{1}{R}-\frac{1}{r}] \end{aligned} $$

Answer Given, surface tension of water

$$ (S)=7.28 \times 10^{-2} N / m $$

Vapour pressure $(p)=2.33 \times 10^{3} Pa$

The drop will evaporate, if the water pressure is greater than the vapour pressure.

Let a water droplet or radius $R$ can be formed without evaporating.

Vapour pressure $=$ Excess pressure in drop.

$$ \begin{matrix} \therefore \quad & p=\frac{2 S}{R} \\ \text { or } \quad R & =\frac{2 S}{p}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^{3}} \\ & =6.25 \times 10^{-5} m \end{matrix} $$

Thinking Process

As we are going up in the atmosphere, thickness of the gases above us decreases hence, pressure also decreases.

Answer (a) Consider a horizontal parcel of air with cross-section $A$ and height $d h$.

Let the pressure on the top surface and bottom surface be $p$ and $p+d p$. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.

$\begin{aligned} & \text { i.e., } \quad(p+d p) A-p A=-\rho g A d h \quad(\because \text { Weight }=\text { Density } \times \text { Volume } \times g) \ & =-\rho \times A d h \times g \ & \Rightarrow \quad d p=-\rho g d h . \quad(\rho=\text { density of air }) \ & \end{aligned}$

Negative sign shows that pressure decreases with height.

(b) Let $p_0$ be the density of air on the surface of the earth.

As per question, pressure $\propto$ density

$$ \begin{aligned} & \Rightarrow \quad \frac{p}{p_0}=\frac{\rho}{\rho_0} \\ & \Rightarrow \quad \rho=\frac{\rho_0}{p_0} p \\ & \therefore \quad d p=-\frac{\rho_{o} g}{p_0} p d h \quad[\because d p=-\rho g d h] \\ & \Rightarrow \quad \frac{d p}{p}=-\frac{\rho_0 g}{p_0} d h \end{aligned} $$

$$\Rightarrow \quad \int_{p _0} ^p \frac{d p}{p}=-\frac{\rho _0 g}{p _0} \int _0 ^h d h \quad \biggl[\begin{array}{l}\because \text { at } h=0, r= p_0 \\ \text { and } \quad \text { at } h=h, p=p\end{array} \biggl]$$

$$ \begin{aligned} & \Rightarrow \quad \ln \frac{p}{p_0}=-\frac{\rho_0 g}{p_0} h \\ & \text { By removing log, } \quad p=p_0 e(-\frac{\rho_0 g h}{p_0}) \end{aligned} $$

(c) As $p=p_0 e^{-\frac{\rho_0 g h}{p_0}}$,

$$ \begin{matrix} \Rightarrow & \ln \frac{p}{p_0} & =-\frac{\rho_0 g h}{P_0} \\ & \text { By question, } & p & =\frac{1}{10} p_0 \\ \Rightarrow & \ln (\frac{\frac{1}{10} p_0}{p_0}) & =-\frac{\rho_0 g}{p_0} h \\ \Rightarrow & \ln \frac{1}{10} & =-\frac{\rho_0 g}{p_0} h \rho_0 \end{matrix} $$

(d) We know that

$$ \begin{aligned} h & =-\frac{p_0}{\rho_{o} g} \ln \frac{1}{10}=-\frac{p_{o}}{p_0 g} \ln (10)^{-1}=\frac{p_{o}}{p_0 g} \ln 10 \\ & =\frac{p_0}{\rho_{o} g} \times 2.303 \quad \quad[\because \ln (x)=2.303 \log _{10}(x)] \\ & =\frac{1.013 \times 10^{5}}{1.22 \times 9.8} \times 2.303=0.16 \times 10^{5} m \\ & =16 \times 10^{3} m \end{aligned} $$

Temperature $(T)$ remains constant only near the surface of the earth, not at greater heights.

Answer (a) Given , $L_{v}=540 kcal kg^{-1}$

$$ =540 \times 10^{3} cal kg^{-1}=540 \times 10^{3} \times 4.2 J kg^{-1} $$

$\because \quad$ Energy required to evaporate $1 kg$ of water $=L_{V} kcal$

$$ \therefore \quad M_{A} kg \text { of water requires } M_{A} L_{V} kcal \quad[\because Q=m L] $$

Since, there are $N_{A}$ molecules in $M_{A} kg$ of water the energy required for 1 molecule to evaporate is

$$ \begin{aligned} U & =\frac{M_{A} L_{V}}{N_{A}} J \quad[\text { where } N_{A}=6 \times 10^{26}=\text { Avogadro number }] \\ & =\frac{18 \times 540 \times 4.2 \times 10^{3}}{6 \times 10^{26}} J \\ & =90 \times 18 \times 4.2 \times 10^{-23} J \\ & =6.8 \times 10^{-20} J \end{aligned} $$

(b) Let the water molecules to be points and are separated at distance $d$ from each other. Volume of $N_{A}$ molecule of water $=\frac{M_{A}}{\rho_{w}}$

$$ [\because V=\frac{M}{\rho}] $$

Thus, the volume around one molecule is $=\frac{M_{A}}{N_{A} \rho_{w}}$

The volume around one molecule is

$$ \begin{aligned} d^{3} & =(M_{A} / N_{A} \rho_{w}) \\ \therefore \quad & d=(\frac{M_{A}}{N_{A} \rho_{w}})^{1 / 3}=(\frac{18}{6 \times 10^{26} \times 10^{3}})^{1 / 3} \\ (30 \times 10^{-30})^{1 / 3} m & \approx 3.1 \times 10^{-10} m \end{aligned} $$

(c) $\because 1 kg$ of vapour occupies volume $=1601 \times 10^{-3} m^{3}$

$\therefore 18 kg$ of vapour occupies $18 \times 1601 \times 10^{-3} m^{3}$

$6 \times 10^{26}$ molecules occupies $18 \times 1601 \times 10^{-3} m^{3}$

$\therefore 1$ molecule occupies $\frac{18 \times 1601 \times 10^{-3}}{6 \times 10^{26}} m^{3}$

If $d$ is the inter- molecular distance, then

$$ \begin{aligned} d_1^{3} & =(3 \times 1601 \times 10^{-29}) m^{3} \\ d_1 & =(30 \times 1601)^{1 / 3} \times 10^{-10} m \\ & =36.3 \times 10^{-10} m \end{aligned} $$

$$ \therefore \quad d_1=(30 \times 1601)^{1 / 3} \times 10^{-10} m $$

(d) Work done to change the distance from $d$ to $d_1$ is $=F(d_1-d)$

This work done is equal to energy required to evaporate 1 molecule.

$$ \begin{aligned} \therefore \quad F(d_1-d) & =6.8 \times 10^{-20} \\ \text { or } \quad F & =\frac{6.8 \times 10^{-20}}{d_1-d} \\ & =\frac{6.8 \times 10^{-20}}{(36.3 \times 10^{-10}-3.1 \times 10^{-10})} \\ & =2.05 \times 10^{-11} N \end{aligned} $$

(e) Surface tension $=\frac{F}{d}=\frac{2.05 \times 10^{-11}}{3.1 \times 10^{-10}}=6.6 \times 10^{-2} N / m$.

Thinking Process

Pressure inside the curved surface will be greater than of outside pressure.

Answer Let the pressure inside the balloon be $p_{i}$ and the outside pressure be $p_0$, then excess pressure is $p_{i}-p_0=\frac{2 S}{r}$. where, $S=$ Surface tension

$r=$ radius of balloon

Considering the air to be an ideal gas $p_{j} V=n_{j} R T_{j}$ where, $V$ is the volume of the air inside the balloon, $n_{i}$ is the number of moles inside and $T_{i}$ is the temperature inside, and $p_0 V=n_0 R T_0$ where $V$ is the volume of the air displaced and $n_0$ is the number of moles displaced and $T_0$ is the temperature outside.

So.

$$ n_{i}=\frac{p_{i} V}{R T_{i}}=\frac{M_{i}}{M_{A}} $$

where, $M_{i}$ is the mass of air inside and $M_{A}$ is the molar mass of air

and

$$ n_0=\frac{p_0 V}{R T_0}=\frac{M_0}{M_{A}} $$

where, $M_{O}$ is the mass of air outside that has been displaced. If $w$ is the load it can raise, then $w+M_{i} g=M_{o} g$

$\Rightarrow \quad W=M_0 g-M_{j} g$

As in atmosphere $21 % O_2$ and $79 % N_2$-is present

$\therefore$ Molar mass of air

$$ M_{i}=0.21 \times 32+0.79 \times 28=28.84 g $$

$\therefore$ Weight raised by the balloon

$$ \begin{aligned} W & =(M_0-M_{i}) g \\ \Rightarrow \quad W & =\frac{M_{A} V}{R}(\frac{p_0}{T_0}-\frac{p_{i}}{T_{i}}) g \\ & =\frac{0.02884 \times \frac{4}{3} \pi \times 8^{3} \times 9.8}{8.314} \quad(\frac{1.013 \times 10^{5}}{293}-\frac{1.013 \times 10^{5}}{333}-\frac{2 \times 5}{8 \times 313}) \\ & =\frac{0.02884 \frac{4}{3} \pi \times 8^{3}}{8.314} \times 1.013 \times 10^{5}(\frac{1}{293}-\frac{1}{333}) \times 9.8 \\ & =3044.2 N \end{aligned} $$

$\therefore \quad$ Mass lifted by the balloon $=\frac{w}{g}=\frac{3044.2}{10} \approx 304.42 kg$.

$$ \approx 305 kg \text {. } $$