Fluid Mechanics

To determine the work done in blowing a soap bubble, we consider the surface tension and the change in the surface area of the bubble.

Given:

Initial radius of the bubble = $ R $

Work done to blow the initial bubble = $ W $

New radius of the bubble = $ 2R $

The formula for the work done in blowing a soap bubble is:

$ W = \text{Surface Tension} \times \text{Change in Area} $

Initial Case (Radius $ R $)

For the initial bubble with radius $ R $, the work done $ W $ is calculated as:

$ W_1 = (2T) \times 4\pi R^2 = 8\pi R^2 T $

New Case (Radius $ 2R $)

For the bubble with a new radius $ 2R $, the work done $ W_2 $ is:

$ W_2 = (2T) \times 4\pi(2R)^2 = 32\pi R^2 T $

Expressing $ W_2 $ in terms of $ W $:

$ W_2 = 4(8\pi R^2 T) = 4W $

Thus, the work done when the bubble is blown to a radius $ 2R $ is $ 4W $.

Three identical vessels labeled A, B, and C are filled with different liquids that have equal masses but varying densities: $\rho_A$, $\rho_B$, and $\rho_C$, with the relationship $\rho_A > \rho_B > \rho_C$.

When the density is lower, the liquid occupies a greater volume, which means it will reach a higher height in the vessel, since the mass is constant across all vessels.

The volume of each liquid is given by:

$ \begin{aligned} \text{Volume}_A &= \frac{m}{\rho_A}, \\ \text{Volume}_B &= \frac{m}{\rho_B}, \\ \text{Volume}_C &= \frac{m}{\rho_C}. \end{aligned} $

Given that the vessels have the same cross-sectional area $a$, the heights of the liquids in the vessels can be calculated as:

$ \begin{aligned} H_A &= \frac{m}{a \rho_A}, \\ H_B &= \frac{m}{a \rho_B}, \\ H_C &= \frac{m}{a \rho_C}. \end{aligned} $

The pressure at the bottom of each vessel, determined by the pressure at a given depth, is calculated using:

$ p = \rho g H $

Substituting for each vessel, we find:

$ \begin{aligned} p_A &= \frac{(\rho_A g) m}{a \rho_A}, \\ p_B &= \frac{(\rho_B g) m}{a \rho_B}, \\ p_C &= \frac{(\rho_C g) m}{a \rho_C}. \end{aligned} $

Simplifying these expressions shows that:

$ p_A = p_B = p_C $

Thus, the pressure at the bottom of the vessels is the same for each liquid, irrespective of the liquid's density.

Given:

Radius of the straw: $ R $

Surface tension of the liquid: $ T $

Contact angle: $ 53^\circ $

$ \cos 53^\circ = 0.6 $

The force due to surface tension is calculated using the formula:

$ F = T \times 2 \times (\text{length} + \text{thickness}) \times \cos \theta $

Since the thickness is negligible, the formula simplifies to:

$ F = T \times 2 \times l \times \cos \theta $

The length $ l $ of the contact line around the straw is the circumference of the straw's circular cross-section, which is $ 2\pi R $.

Thus, the force $ F $ becomes:

$ \begin{aligned} F &= T \times 2 \times 2\pi R \times \cos 53^\circ \\ &= T \times 4\pi R \times 0.6 \\ &= 2.4\pi R T \\ &= \frac{12\pi R T}{5} \end{aligned} $

Hence, the force acting on the straw due to the surface tension of the liquid is $\frac{12\pi R T}{5}$.

When a solid metal sphere is released into a vertical liquid column and reaches terminal velocity, there is a balance of forces in play. The terminal velocity is the highest velocity achieved by an object as it falls through a fluid, at which point the net force acting on it is zero, and it stops accelerating.

In this scenario:

$F_B$: Buoyant force, acts upward.

$F_V$: Viscous force, acts upward, resisting motion.

$F_W$: Gravitational force, acts downward.

Since the object has reached terminal velocity, the sum of the upward forces (buoyant and viscous) equals the downward gravitational force. Hence, the correct relationship between these forces is:

$ F_W = F_B + F_V $

This equation indicates that the gravitational force acting downward is balanced by the buoyant force and the viscous force, both acting upward.

To determine the shear stress between layers of water at $y=a$, we start by using the given velocity profile and the definition of shear stress for a fluid in laminar flow.

The shear stress $\tau$ in a fluid is given by:

$ \tau = \eta \frac{du}{dy} $

where $\eta$ is the dynamic viscosity of the fluid, and $\frac{du}{dy}$ is the velocity gradient in the direction perpendicular to the flow.

Given the velocity profile:

$ v = k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) $

First, we need to calculate the velocity gradient $\frac{dv}{dy}$:

$ \frac{dv}{dy} = \frac{d}{dy} \left( k \left( \frac{2 y^2}{a^2} - \frac{y^3}{a^3} \right) \right) $

Applying the differentiation, we get:

$ \frac{dv}{dy} = k \left( \frac{d}{dy} \left( \frac{2 y^2}{a^2} \right) - \frac{d}{dy} \left( \frac{y^3}{a^3} \right) \right) $

Differentiating term by term, we have:

$ \frac{dv}{dy} = k \left( \frac{2 \cdot 2y}{a^2} - \frac{3y^2}{a^3} \right) = k \left( \frac{4y}{a^2} - \frac{3y^2}{a^3} \right) $

We need to evaluate this gradient at $y = a$:

$ \left. \frac{dv}{dy} \right|_{y=a} = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4a}{a^2} - \frac{3a^2}{a^3} \right) = k \left( \frac{4}{a} - \frac{3}{a} \right) = k \left( \frac{4-3}{a} \right) = \frac{k}{a} $

Now, using the given viscosity of the water, $2 \eta$, the shear stress $\tau$ is given by:

$ \tau = 2 \eta \left( \frac{k}{a} \right) = \frac{2 \eta k}{a} $

Therefore, the shear stress between layers of water at $y=a$ is:

$ \tau = \frac{2 \eta k}{a} $

The correct option is Option A:

$ \frac{2 \eta k}{a} $

Let, $H=$ height of liquid in vessel and $h=$ height of hole.

Velocity of outflow, $v=\sqrt{2 g h}$

Time taken by liquid stream to reach bottom is

$ t=\sqrt{\frac{2(H-h)}{g}} $

Range of liquid $x$ is

$ \begin{aligned} x & =v \cdot t \\ & =\sqrt{2 g h} \times \sqrt{\frac{2(H-h)}{g}} \\ & =2 \sqrt{h(H-h)} \end{aligned} $

For maximum range,

$ \frac{d x}{d h}=0 \Rightarrow h=\frac{H}{2} $

So, range $x$ will be maximum when

$ h=\frac{H}{2} $

Hence, in given case, $h=\frac{50}{2}=25 \mathrm{~cm}$

Let $r=$ radius of large drop and $R=$ radius of each of small drops

As volume remains same, Initial volume $=$ Final volume

$ \begin{aligned} & \frac{4}{3} \pi r^3 =8 \times \frac{4}{3} \pi R^3 \\ \Rightarrow & r =2 R \text { or } R=\frac{r}{2} \end{aligned} $

Change in surface area due to splitting of drop is

$ \begin{aligned} \Delta A & =8 \times 4 \pi R^2-4 \pi r^2 \\ & =8 \times 4 \pi\left(\frac{r}{2}\right)^2-4 \pi r^2 \\ & =4 \pi r^2 \end{aligned} $

∴ Work done $=$ S. $\Delta A=S \cdot 4 \pi r^2=4 \pi r^2 S$

For a horizontal venturimeter by Bernoulli's

theorem we have,

$ \begin{aligned} p_1+\frac{1}{2} \rho v_1^2 & =p_2+\frac{1}{2} \rho v_2^2 \\ \Rightarrow \quad p_1-p_2 & =\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \end{aligned} $

Also, $A_1 v_1=A_2 v_2$

(continuity equation)

So we have,

$ p_1-p_2=\frac{1}{2} \rho\left[\left(\frac{A_1 v_1}{A_1}\right)^2-v_1^2\right] $

Here, $\rho=1000 \mathrm{kgm}^{-3}$

$ \begin{aligned} \frac{A_1}{A_2} & =\frac{\pi d_1^2}{\pi d_2^2}=\left(\frac{d_1}{d_2}\right)^2=\left(\frac{4}{2}\right)^2=4 \\ v_1 & =10 \mathrm{~m} / \mathrm{s} \end{aligned} $

So, from Eq. (i), we get

$ \begin{aligned} p_2-p_1 & =\frac{1000}{2} \times(4 \times 100-100) \\ & =500 \times 300=1.5 \times 10^5 \mathrm{~Pa} \end{aligned} $

Surface energy or work done to increase size of bubble = Change in surface area × Surface tension

$ =4 \pi\left[(2 R)^2-R^2\right] \times T \times 2=24 \pi R^2 T $

[Note-Factor 2 appears here because a bubble has two surfaces.]

We know that, blood flow rate through an artery

$ \begin{aligned} & v_c=\frac{R_c \eta}{2 \rho r} \\ & \therefore \quad v_c \propto \frac{\eta}{\rho r} \\ & \therefore \frac{\Delta v_c}{v_c} \times 100=\frac{\Delta \eta}{\eta} \times 100+\frac{\Delta \rho}{\rho} \times 100 +\frac{\Delta r}{r} \times 100 \\ & \quad=1 \%+1 \%+1 \%=3 \% \end{aligned} $

Length of metal cube,

$ 1=10 \mathrm{~cm} $

∴ Area of surface, $A=l^2=10^2 \mathrm{~cm}^2$

$ =100 \mathrm{~cm}^2=10^{-2} \mathrm{~m}^2 $

Thickness of film, $\Delta x=0.2 \mathrm{~mm}$

$ \begin{gathered} =2 \times 10^{-4} \mathrm{~m} \\ F=0.1 \mathrm{~N} \\ \Delta y=0.08 \mathrm{~m} / \mathrm{s}=8 \times 10^{-2} \mathrm{~m} / \mathrm{s} \end{gathered} $

We know that, viscous force

$ \begin{array}{rlrl} F & =\eta \frac{\Delta v}{\Delta x} \cdot A \\ \Rightarrow & 0.1 =\eta \times \frac{8 \times 10^{-2}}{2 \times 10^{-4}} \times 10^{-2} \\ \Rightarrow & 0.1 =\eta \times 4 \\ \Rightarrow & \eta =\frac{0.1}{4}=0.025 \\ & =25 \times 10^{-2} \mathrm{~N}-\mathrm{s} / \mathrm{m}^2 \end{array} $

Area of piston $=\frac{n \cdot x \cdot g}{A}$

Area of piston $=\frac{(n-1) \cdot x \cdot g}{A}$

So, $\quad \frac{p-p^{\prime}}{p}=\frac{[n-(n-1)] \times g / A}{n \times g / A}=\frac{1}{n}$.

The given situation is shown below

$ \begin{aligned} &\text { According to principle of continuity, }\\ &\begin{aligned} A_1 v_1 & =A_2 v_2 \\ v_1 & =\frac{A_2 v_2}{A_1}=\frac{\pi R_2^2 \times 4 v}{\pi R_1^2} \\ & =\frac{\pi R^2 \times 4 v}{\pi(2 R)^2}=\frac{4 \pi R^2 v}{4 \pi R^2}=v \end{aligned} \end{aligned} $

The given situation is shown below.

Let the desity of material of sphere $=\rho^{\prime}$ Weight of hollow sphere = Buoyant force

$ \begin{aligned} & \Rightarrow \quad \rho^{\prime} \frac{4}{3} \pi\left[4^3-2^3\right] g \\ & \quad=\text { weight of displaced liquid } \\ & \Rightarrow \quad \rho^{\prime} \frac{4}{3} \pi[64-8] g=\rho \times \frac{1}{2} \times \frac{4}{3} \pi \times 4^3 \times g \\ & \Rightarrow \quad \rho^{\prime} \times 56=\rho \times \frac{1}{2} \times 64 \\ & \Rightarrow \quad \rho^{\prime}=2 \times \frac{1}{2} \times \frac{64}{56}=1.14 \mathrm{~g} / \mathrm{cm}^3 \end{aligned} $

We know that,

$ v_{\text {Terminal }}=\frac{2 g r^2}{9 \eta}\left(\rho_b-\rho_l\right) $

where, $\rho_b=$ density of body

$ =1000 \mathrm{~kg} / \mathrm{m}^3 $

$\rho_l=$ density of liquid or air $=0 \mathrm{~kg} / \mathrm{m}^3$

$ \begin{aligned} v_{\text {Terminal }} & =\frac{2}{9} \times \frac{10}{1.8 \times 10^{-5}} \times\left(\frac{0.02}{1000}\right)^2 \\ (1000-0) & \\ v_{\text {Terminal }} & =4.9 \mathrm{~cm} / \mathrm{s} \end{aligned} $

The situation given in the question is depicted below

$ v=\sqrt{2 \times 10 \times 20}=\sqrt{400}=20 \mathrm{~m} / \mathrm{s} $

Given, rate of flow $=$ area × velocity

$ \begin{aligned} & A v=3.08 \times 10^{-5} \mathrm{~m}^3 / \mathrm{s} \\ \Rightarrow & A=\frac{3.08 \times 10^{-5}}{20} \mathrm{~m}^2 \end{aligned} $

Now, area of circular hole

$ =\pi R^2=\pi\left(\frac{D}{2}\right)^2 $

where, $D=$ diameter of the circle.

$ \begin{aligned} & \Rightarrow \quad \pi\left(\frac{D}{2}\right)^2=\frac{3.08 \times 10^{-5}}{20} \mathrm{~m}^2 \\ & \Rightarrow \quad D^2=\frac{3.08 \times 10^{-5} \times 4}{20 \pi} \\ & \Rightarrow D=\sqrt{\frac{3.08 \times 10^{-5} \times 4}{20 \pi}} \\ & =1.4 \times 10^{-3} \mathrm{~m} \\ & =1.4 \mathrm{~mm} \end{aligned} $

The given situation is shown below

where, $p_0=$ atmospheric pressure,

$R=$ Air bubble of radius $=1 \mathrm{~mm}$,

$\rho=$ density of liquid $=2000 \mathrm{~kg} / \mathrm{m}^3$

and $T=$ surface tension of liquid

$ =0.1 \mathrm{~N} / \mathrm{m} . $

Now, we know that, excess pressure inside an air bubble inside any liquid $=\frac{2 T}{R}$

Also, pressure inside the air bubble $=$ pressure outside + excess pressure

Difference of pressure inside the air bubble and atomospheric pressure,

$ \begin{aligned} & \Rightarrow p_{\text {inside }}-p_0=\rho g h+\frac{2 T}{R} \\ & =2000 \times 10 \times \frac{8}{100}+\frac{2 \times 0.1}{1 \times 10^{-3}} \\ & =1600+(2 \times 100)=1800 \mathrm{~N} / \mathrm{m}^2 \end{aligned} $

Given,

$\begin{aligned} & r_A=10 \mathrm{~cm}=0.1 \mathrm{~m} \\ & r_B=100 \mathrm{~m} \\ & r_C=5 \mathrm{~cm}=0.05 \mathrm{~m} \end{aligned}$

weight placed on piston $A=m \times g$

$=2 \mathrm{~g}$

Using Pascal's law;

$\begin{gathered} \frac{F_A}{A_A}=\frac{F_B}{A_B}=\frac{F_C}{A_C} \\ \Rightarrow \quad \frac{2 g}{\pi(0.1)^2}=\frac{F_B}{\pi(100)^2}=\frac{F_C}{\pi(0.05)^2} \quad \text{... (i)} \end{gathered}$

From Eq. (i) we have

$\begin{aligned} F_B & =\frac{2 g}{\pi(0.1)^2} \times \frac{\pi(100)^2}{1}=\frac{2 \times 10000 \times g}{0.01} \\ & =2 \times 10^6 \mathrm{~g} \end{aligned}$

Similarly, $F_{\mathrm{c}}=\frac{\pi(0.05)^2}{1} \times \frac{2 \mathrm{~g}}{\pi(0.1)^2}$

$=\frac{0.0025}{0.01} \times 2 \mathrm{~g}=0.50 \mathrm{~g}$

Hence, $B$ can lift mass of $2 \times 10^6 \mathrm{~kg}$ and $C$ can lift mass of 0.50 kg.

Therefore, no option is correct.

In a hydraulic lift, compressed air applies a force $F$ on a small piston with a radius of 3 cm. This pressure causes a second piston, which has a radius of 5 cm, to lift a load weighing 1875 kg. We want to determine the value of $F$, considering the acceleration due to gravity is $10 \mathrm{~ms}^{-2}$.

Using the principle of hydraulic lifts, we have:

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

This can be rewritten based on the areas of the pistons:

$\begin{aligned} &\Rightarrow \quad \frac{F}{\pi r_1^2} = \frac{mg}{\pi r_2^2} \\ &\Rightarrow \quad \frac{F}{r_1^2} = \frac{mg}{r_2^2} \\ &\Rightarrow \quad \frac{F}{(3 \times 10^{-2})^2} = \frac{1875 \times 10}{(5 \times 10^{-2})^2} \\ &\Rightarrow \quad F = 18750 \times \frac{9 \times 10^{-4}}{25 \times 10^{-4}} \\ &\Rightarrow \quad F = 6750 \mathrm{~N} \\ \end{aligned}$

For the first limb of the tube, height ascent by liquid

$h_1=\frac{2 T \cos \theta}{r_1 \rho g}$

$\begin{aligned} \text { where, } T & =\text { surface tension }, \\ \theta & =\text { angle of contact } \\ r_1 & =\text { radius of tube, } \\ \rho & =\text { density of liquid and } \\ g & =\text { gravitational acceleration. } \end{aligned}$

For the second limb of the tube, height ascent by liquid,

$\begin{aligned} h_2 & =\frac{2 T \cos \theta}{r_2 \rho g} \\ \therefore h_1-h_2 & =\frac{2 T \cos \theta}{\rho g}\left[\frac{1}{r_1}-\frac{1}{r_2}\right] \end{aligned}$

$\begin{aligned} \text { Given, } T & =0.03 \mathrm{Nm}^{-1} \\ r_1 & =2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}, \\ r_2 & =4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}, \\ \theta & =0^{\circ}, \rho=150 \mathrm{~kg} \mathrm{~m}^{-3} \text { and } g=10 \mathrm{~ms}^{-2} \\ h_1-h_2 & =\frac{2 \times 0.03 \times \cos 0^{\circ}}{1500 \times 10}\left[\frac{1}{2 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}}\right] \\ & =0.001 \mathrm{~m}=1 \mathrm{~mm} \end{aligned}$

The given situation is shown below

$h=h_1-h_2=10 \mathrm{~cm}=0.1 \mathrm{~m}$

By the principle of continuity,

$\begin{aligned} & & A_1 v_1 & =A_2 v_2 \\ \Rightarrow & & 2 A \times \sqrt{2} & =A \times v_2 \\ \Rightarrow & & v_2 & =2 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

According to Bernoulli's theorem,

$\begin{aligned} & p_1+\frac{1}{2} \rho v_1^2+\rho g h_1=p_2+\frac{1}{2} \rho v_2^2+\rho g h_2 \\ & \Rightarrow\left(p_1-p_2\right)+\rho g\left(h_1-h_2\right)=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\ & \Rightarrow \quad 100+\rho g(0.1)=\frac{1}{2} \rho\left[(2 \sqrt{2})^2-(\sqrt{2})^2\right] \\ & \Rightarrow \quad 100+\rho \times 10 \times 0.1=\frac{1}{2} \rho(8-2) \\ & \Rightarrow \quad 100+\rho=3 \rho \\ & \Rightarrow \quad 100=2 \rho \\ & \Rightarrow \quad \rho=50 \mathrm{~kg} \mathrm{~m}^{-3} \\ & \end{aligned}$

With the increase of temperature, the viscosity of gases increases due to increase of their thermal velocity whereas, viscosity of liquids decreases with the increse of temperature. Water does not wet an oily glass because cohesive force between water molecules is less than adhesive force. between the molecules of oil and glass.

A liquid will wet a surface of a solid if the angle of contact is acute (i.e less than $90^{\circ}$).

Hence, statement $(\mathrm{A})$ is true and statement (B) and (C) are false.

Liquid volume coming out of the tube per second

$V = {{p\pi {r^4}} \over {8\eta l}}$

$ \Rightarrow V\,\alpha {r^4}$

$ \Rightarrow {{{V_2}} \over {{V_1}}} = {\left[ {{{{r_2}} \over {{r_1}}}} \right]^4} \Rightarrow {V_2} = {V_1}{\left[ {{{125} \over {100}}} \right]^4} = {V_1}{\left( {{5 \over 4}} \right)^4}$

$ \Rightarrow {V_2} = {V_1}(244)$

So, now ${{\Delta V} \over V} = {{{V_2} - {V_1}} \over {{V_1}}} = {{(244){V_1} - {V_1}} \over {{V_1}}} = 144\% $

As we know that,

Surface tension is that property of any fluid due to which it resist any external force due to cohesive nature of fluid that cause liquid to have minimum area.

Since, aeroplane works on Bernoulli’s principle in which fast moving air above wing exert less pressure than air below it which pushes the wing in upward direction, so wings are made convex from upper side and concave from lower side. Hence, Assertion is true but Reason is false.

Given,

Weight of glass, $W=390 \mathrm{~g}$

internal volume, $V_i=500 \mathrm{~cc}$

Let specific gravity of glass $=\rho$

Density of water $=1 \mathrm{~gcc}^{-1}$

As we know that,

$\begin{aligned} & \text { Density }(d)=\frac{\text { Mass }(m)}{\text { Volume }(V)} \\ & \Rightarrow \quad m=d V \Rightarrow\left(390+\frac{500}{2}\right)=1 \times V \\ & \Rightarrow \quad(390+250)=V \Rightarrow V=640 \mathrm{~cc} \\ & \therefore \text { Volume of glass, } V_g=V-V_{\text {in }} \\ & =640-500=140 \mathrm{~cc} \\ & \therefore \rho=\frac{390}{140}=2.78 \mathrm{~gcc}^{-1}=2.8 \mathrm{~gcc}^{-1} \\ & \end{aligned}$

Cohesive force is defined as the force of attraction between intermolecular particles. Since, cohesive force between the molecules of water is greater than adhesive force between water and oil molecules.

$\therefore$ Water does not wet glass.

As we know that,

Reynold's number $\left(R_e\right)$ is used to determine the nature of flow of water.

If $R_e$ lies between 0 to 2000, then flow is laminar and steady.

For $200 \leq R_e < 300$. flow is unstable.

And for $R_e > 3000$, flow is turbulent.

$\therefore$ Option (c) is incorrect.

Given, capillary rise, $H=7.5 \mathrm{~cm}=7.5 \times 10^{-2} \mathrm{~m}$

Surface tension, $T=7.5 \times 10^{-2} \mathrm{Nm}^{-1}$

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

Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$

Let radius of capillary tube be $R$.

Density of water, $\rho=1000 \mathrm{~kg} / \mathrm{m}^3$

Since, $H =\frac{2 T \cos \theta}{\rho g R}$

$\Rightarrow \quad R =\frac{2 T \cos \theta}{\rho g H}=\frac{2 \times 7.5 \times 10^{-2} \times \cos 0^{\circ}}{1000 \times 10 \times 7.5 \times 10^{-2}}$

$=2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm}$

Given, horizontal tube of variable diameter By using equation of continuity

$A v=\text { constant }$ .... (i)

where, $A$ is area of cross-section and $v$ is velocity of fluid through that cross-section.

As, $p=F / A \Rightarrow A=F / p$

Substituting in Eq. (i), we get

$F / p \cdot v=\text { constant }$

If force is same.

$\therefore \quad \frac{v}{p}=\text { constant } $

If pressure is least, then velocity is highest.

Volume of bigger drop = Volume of two smaller drops

$ \Rightarrow {4 \over 3}\pi R{'^3} = 2 \times {4 \over 3}\pi {R^3}$

where, R' is the radius of bigger drop.

$ \Rightarrow R{'^3} = 2{R^3}$

$ \Rightarrow R' = {2^{1/3}}\,R$

Initial surface energy of two small drops, ${U_1} = 8\pi {R^2}\sigma $ ..... (i)

Final surface energy of big drop, ${U_2} = 4\pi R{'^2}\sigma $

$ = 4\pi \times {2^{2/3}}{R^2}\sigma $ ..... (ii)

On dividing Eq. (ii) by Eq. (i), we get

${{{U_2}} \over {{U_1}}} = {{4\pi \times {2^{2/3}}\,{R^2}\sigma } \over {8\pi {R^2}\sigma }} = {{{2^{2/3}}} \over 2} = {{{2^{ - 1/3}}} \over 1}$

Ratio of surface energy of bigger drop to smaller one is 2^$-$1/3 : 1.

According to given situation, Volume of big drop $=$ Volume of $n$ small drops

$ \begin{align*} \frac{4}{3} \pi R^3 & =n \times \frac{4}{3} \pi r^3 \\ \Rightarrow \quad r^3 & =\frac{R^3}{n} \text { or } r=\frac{R}{n^{1 / 3}} \\ n r^2 & =\frac{n R^2}{n^{2 / 3}}=n^{1 / 3} R^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Surface area of big drop, $A=4 \pi R^2$ Surface area of $n$ small drops,

$ \begin{align*} A^{\prime} & =n \times 4 \pi r^2 \\ \Delta A & =A^{\prime}-A=4 \pi n r^2-4 \pi R^2 \\ & =4 \pi\left(n r^2-R^2\right) \\ & =4 \pi\left(n^{1 / 3} R^2-R^2\right) \text { [from Eq. } \\ & =4 \pi R^2\left(n^{1 / 3}-1\right) \end{align*} $

The work done for change in area,

$ \begin{aligned} W & =T \times \Delta A \\ & =T \times 4 \pi R^2\left(n^{1 / 3}-1\right) \end{aligned} $

This work is store in the form of energy.

$ \text { According to Pascal's law, } P_1=P_2 $

$ \begin{array}{ll} \Rightarrow & \frac{F_1}{A_1}=\frac{F_2}{A_2} \\ \Rightarrow & \frac{1000 \mathrm{~g}}{1}=\frac{m_2 \mathrm{~g}}{0.01} \\ & {\left[\because A_1=1 \mathrm{~m}^2, A_2=0.01 \mathrm{~m}^2\right]} \\ \Rightarrow & m_2=1000 \times 0.01=10 \mathrm{~kg} \end{array} $

$ \begin{aligned} & \text {Volume of wood block }=\frac{\text { Mass }}{\text { Density }} \\ & =\frac{160}{0.8 \times 1}=\frac{1600}{8}=200 \mathrm{~cm}^3 \end{aligned} $

Let mass of metal piece is $x \mathrm{~g}$.

$ \begin{gathered} \text { Volume of metal piece }=\frac{\text { Mass }}{\text { Density }} \\ =\frac{x}{10 \times 1}=\frac{x}{10} \mathrm{~cm}^3 \end{gathered} $

Total volume of fluid displaced

$ =\left(200+\frac{x}{10}\right) \times 10^{-6} \mathrm{~m}^3 $

Weight of fluid displaced $=V \times \rho_f \times g$

$ \begin{aligned} & =\left(200+\frac{x}{10}\right) 10^{-6} \times 1000 \times 10 \\ & =\left(200+\frac{x}{10}\right) 10^{-2} \mathrm{~N} \end{aligned} $

If block floats in water, then weight $=$ upthrust

$ \begin{gathered} \Rightarrow(160+x) 10 \times 10^{-3}=\left(200+\frac{x}{10}\right) \times 10^{-2} \\ 160+x=200+\frac{x}{10} \Rightarrow \frac{9 x}{10}=40 \\ x=\frac{400}{9}=44.4 \mathrm{~g} \end{gathered} $

Radius of meniscus, $R=1 \mathrm{~cm}$

                                                      $ =10^{-2} \mathrm{~m} $

Number of smaller droplets, $n=10^6$

Surface tension, $T=435 \times 10^{-3} \mathrm{Nm}^{-1}$

Work done in splitting a large drop of radius $R$ in $n$ smaller drops is

$ W=T \times \Delta A $

where, $\Delta A=$ increase in area

and $\quad T=$ surface tension.

$ \begin{aligned} \therefore \quad W & =\left(4 \pi R^2\right) T \cdot\left(n^{1 / 3}-1\right) \\ = & 4 \times \pi \times\left(1 \times 10^{-2}\right)^2 \times 435 \times 10^{-3} \times\left(\left(10^6\right)^{1 / 3}-1\right) \\ & \quad=54.1 \times 10^{-3} \mathrm{~J} \end{aligned} $