Mechanical Properties of Fluids

Surface tension is property of surface molecules not interior molecules. In case of liquid as temperature increases surface tension decreases. In case of gases due to increase in temperature collision will be more frequent hence viscosity increases.

Given:

$A_1=1.0c{m}^2 , A_2=20m{m}^2=0.2c{m}^2$

Using continuity equation:

$A_1{v}_1=A_2{v}_2$

$1\times 10=0.2\times v_2\Rightarrow v_2=50cm/s$

Using Bernoulli’s theorem at the same horizontal level:

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

$P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$

Convert to SI:

$v_1=0.1m/s,v_2=0.5m/s$

$P_1-P_2=\frac{1}{2}\left(600\right)\left(0.5^2-0.1^2\right)=72Pa$

Here, P_i - P_o = 2100 ... (1)

Let the pressure at the depth is P_o'.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{o}}^{\text{'}} = {\mathrm{P}}_{\mathrm{o}} + \mathrm{\rho } \mathrm{g} \mathrm{h} \\ \mathrm{So}, \\ {\mathrm{P}}_{\mathrm{i}} -{\mathrm{P}}_{\mathrm{o}}^{\text{'}} = \frac{2\mathrm{T}}{\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{i}} ={\mathrm{P}}_{\mathrm{o}}+ \mathrm{\rho } \mathrm{g} \mathrm{h} + \frac{2\mathrm{T}}{\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{o}} + 2100 = {\mathrm{P}}_{\mathrm{o}} + 1000\times 10\times 0.2 + \frac{2\mathrm{T}}{{10}^{-3}} \end{gathered}$$

$100=\frac{2T}{R}$

$\Rightarrow T=\frac{100\times {10}^{-2}}{20}$

= 0.05 N/m

For steady rise, viscous force balances buoyant force:

$6\pi \eta rv=\frac{4}{3}\pi r^3\rho g$

$\eta =\frac{2r^2\rho g}{9v}$

$r=1mm={10}^{-3}m,v=0.5cm/s= 5 \times {10}^{–3} m/s$

$\eta =\frac{2{\left({10}^{-3}\right)}^2\times 2000\times 10}{9·5\times {10}^{-3}}=\frac{2\times {10}^{-6}\times 20000}{45\times {10}^{-3}}$

$=\frac{0.04}{0.045}\approx 0.88Pa=8.8$ poise

$$\begin{gathered} \text{ Apply Bernoulli's eqn } \\ \frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g} \\ \frac{P_1}{\rho g}+0=\frac{P_2}{\rho g}+\frac{V^2}{2g} \\ V=\sqrt{\frac{2\left(P_1-P_2\right)}{\rho }} \\ V\propto \sqrt{\left(P_1-P_2\right)} \end{gathered}$$

Here, P_i - P_o = 2100 ... (1)

Let the pressure at the depth is P_o'.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{o}}^{\text{'}} = {\mathrm{P}}_{\mathrm{o}} + \mathrm{\rho } \mathrm{g} \mathrm{h} \\ \mathrm{So}, \\ {\mathrm{P}}_{\mathrm{i}} -{\mathrm{P}}_{\mathrm{o}}^{\text{'}} = \frac{2\mathrm{T}}{\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{i}} ={\mathrm{P}}_{\mathrm{o}}+ \mathrm{\rho } \mathrm{g} \mathrm{h} + \frac{2\mathrm{T}}{\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{o}} + 2100 = {\mathrm{P}}_{\mathrm{o}} + 1000\times 10\times 0.2 + \frac{2\mathrm{T}}{{10}^{-3}} \end{gathered}$$

$100=\frac{2T}{R}$

$\Rightarrow T=\frac{100\times {10}^{-2}}{20}$

= 0.05 N/m

At terminal velocity (constant velocity), the net force acting on the sphere is zero. The forces acting on the sphere are:

1. Gravitational Force ($F_g$): Downward force due to the weight of the sphere. $$F_g = Mg$$
2. Buoyant Force ($F_b$​): Upward force due to the displaced glycerin. $$F_b = \rho_{\text{glycerin}} V g$$ where $\rho_{\text{glycerin}}$is the density of glycerin, $V$ is the volume of the sphere, and $g$ is the acceleration due to gravity.
3. Viscous Force ($F_v$​): Upward force due to the viscosity of glycerin. This balances the remaining downward force at terminal velocity.

From equilibrium at terminal velocity:

$$F_v = F_g - F_b$$​

The density of glycerin ($\rho_{\text{glycerin}}$) is half the density of the sphere ($\rho_{\text{sphere}}$):

$$\rho_{\text{glycerin}} = \frac{\rho_{\text{sphere}}}{2}$$​​

The mass $M$ of the sphere is related to its density and volume:

$$M = \rho_{\text{sphere}} V$$

Substitute $\rho_{\text{glycerin}} = \frac{\rho_{\text{sphere}}}{2}$​​ into $F_b$:

$$F_b = \frac{\rho_{\text{sphere}}}{2} V g = \frac{M}{2} g$$

Using $F_v = F_g - F_b$:

$$F_v = Mg - \frac{M}{2}g$$

Simplify:

$$F_v = \frac{M}{2} g$$

The viscous force acting on the sphere is:

$$\boxed{\frac{Mg}{2}}$$​​

This question is based on Bernoulli's equation, which states:

Since the pipe is horizontal ($h$h is constant), the equation simplifies to:

For points A and B:

Rearranging:

Divide through by $\rho/2$:

The correct option is:

Pressure in the air bubble:

$P_{inside}=P_0+\rho gh+\frac{2T}{R}$

Given:

$P_{inside}-P_0=2100$

Hydrostatic pressure:

$$\begin{gathered} P_{\text{inside }}-P_0=\rho gh+\frac{2T}{R} \\ 2100=1000\times 10\times 0.2+\frac{2T}{0.001} \\ 2100=2000+\frac{2T}{0.001} \\ T=\frac{1}{20}=0.05 \end{gathered}$$

Thus, the correct answer is 0.05 .

The equation for the time period T of the oscillations is given as:

$T=2\pi \sqrt{\frac{m}{L^2\rho g}}$

Using

$$\begin{gathered} a^{2}xrg=m{a}_{\text{net }} \\ \frac{L^2\rho g}{m}x=a_{\text{net }} \\ m=10g,L=10cm,\rho =1000kg/m^3 \end{gathered}$$

and putting values in equation

$$\begin{gathered} T=2\pi \times {10}^{-2}S \\ y=2 \end{gathered}$$

Terminal velocity of the ball through a viscous medium

\(v=\frac{2}{9} \times \frac{g}{\eta}(\rho-\sigma) r^2\)

Or \(v=\frac{2}{9} \times \frac{g}{\eta}(\rho) r^2\)
Because for viscous medium of negligible density

$$
\begin{aligned}
& (\sigma=0) \therefore \\
& v=\frac{2}{9} \times \frac{g}{\eta} \times \frac{m}{\frac{4}{3} \pi r^3} \times r^2 \\
& {\left[\because \rho=\frac{m}{\frac{4}{3} \pi r^3}\right]}
\end{aligned}
$$

Or \(v=\frac{2}{9} \times \frac{g}{\eta} \times \frac{m}{\frac{4}{3} \pi r}\)

\(\Rightarrow v \propto \frac{1}{r}\)

For the second ball \(v \propto \frac{1}{2 r}\)

\(\therefore \frac{v}{v}=\frac{2 r}{r} \text { Or } v=\frac{v}{2}\)

Terminal velocity of the ball through a viscous medium

\(v=\frac{2}{9} \times \frac{g}{\eta}(\rho-\sigma) r^2\)

Or \(v=\frac{2}{9} \times \frac{g}{\eta}(\rho) r^2\)
Because for viscous medium of negligible density

$$
\begin{aligned}
& (\sigma=0) \therefore \\
& v=\frac{2}{9} \times \frac{g}{\eta} \times \frac{m}{\frac{4}{3} \pi r^3} \times r^2 \\
& {\left[\because \rho=\frac{m}{\frac{4}{3} \pi r^3}\right]}
\end{aligned}
$$

Or \(v=\frac{2}{9} \times \frac{g}{\eta} \times \frac{m}{\frac{4}{3} \pi r}\)

\(\Rightarrow v \propto \frac{1}{r}\)

For the second ball \(v \propto \frac{1}{2 r}\)

\(\therefore \frac{v}{v}=\frac{2 r}{r} \text { Or } v=\frac{v}{2}\)

The work done is proportional to the change in surface area. The surface area of a sphere is $4 \pi r^2$.

1. Initial bubble area:
   The radius of the bubble is $R$. Its surface area is:

   $$A_{\text{initial}} = 4 \pi R^2$$

2. For 27 drops:

   • Each drop has radius $r_1$ such that the volume of the bubble equals the total volume of the drops: $$\frac{4}{3} \pi R^3 = 27 \cdot \frac{4}{3} \pi r_1^3 \implies r_1 = \frac{R}{3}$$​
   • Total surface area of 27 drops: $$A_{\text{27 drops}} = 27 \cdot 4 \pi r_1^2 = 27 \cdot 4 \pi \left( \frac{R}{3} \right)^2 = 12 \pi R^2$$
   • Change in surface area: $$\Delta A = A_{\text{27 drops}} - A_{\text{initial}} = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2$$
   • Work done is proportional to $\Delta A$: $$W_1 = k \Delta A \implies k = \frac{10}{8 \pi R^2}$$​

3. For 64 drops:

   • Each drop has radius $r_2$​ such that: $$\frac{4}{3} \pi R^3 = 64 \cdot \frac{4}{3} \pi r_2^3 \implies r_2 = \frac{R}{4}$$​
   • Total surface area of 64 drops: $$A_{\text{64 drops}} = 64 \cdot 4 \pi r_2^2 = 64 \cdot 4 \pi \left( \frac{R}{4} \right)^2 = 16 \pi R^2$$
   • Change in surface area: $$\Delta A = A_{\text{64 drops}} - A_{\text{initial}} = 16 \pi R^2 - 4 \pi R^2 = 12 \pi R^2$$
   • Work done: $$W_2 = k \Delta A = \frac{10}{8 \pi R^2} \cdot 12 \pi R^2 = 15 \, \text{J}$$

Thus, the work done when splitting into 64 drops is 15 J.

Work done = increase in surface energy of soap bubble:

$W=2T\times \Delta A$

Change in area:

$\Delta A=4\pi \left(r_2^2-r_1^2\right)$

$W=2T·4\pi \left(r_2^2-r_1^2\right)=8\pi T\left(r_2^2-r_1^2\right)$

$=8\pi \times 0.03\left[(0.03{)}^2-(0.01{)}^2\right]$

$=1.92\pi \times {10}^{-4}J$

$\alpha =1.92$

There are two liquid-air surfaces in bubble so

$\Delta P=2\left(\frac{2S}{R}\right)=\frac{4S}{R}$

By volume conservation:

$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$

$r=\frac{R}{4}$

For spherical drops falling through a viscous medium,

$v_t\propto r^2$

So,

$\frac{v_1}{v_2}=\frac{R^2}{(R/4{)}^2}=16$

Using volume conservation:
$8\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3\Rightarrow R=2r$
Initial surface energy:

$U_i=8\left(4\pi r^{2}S\right)=32\pi r^{2}S$

Final surface energy:
$U_f=4\pi R^{2}S=4\pi (2r{)}^{2}S=16\pi r^{2}S$
Energy released:
$U_i-U_f=32\pi r^{2}S-16\pi r^{2}S=16\pi r^{2}S$

$\sigma =$ Density of ball

$$\begin{gathered} V_T=\frac{2R^2}{2\eta }(\sigma -\rho )g \\ {V}_T\propto R^2 \end{gathered}$$

As temperature changes Co-efficient of viscosity changes.

At terminal velocity (constant velocity), the net force acting on the sphere is zero. The forces acting on the sphere are:

1. Gravitational Force ($F_g$): Downward force due to the weight of the sphere. $$F_g = Mg$$
2. Buoyant Force ($F_b$​): Upward force due to the displaced glycerin. $$F_b = \rho_{\text{glycerin}} V g$$ where $\rho_{\text{glycerin}}$is the density of glycerin, $V$ is the volume of the sphere, and $g$ is the acceleration due to gravity.
3. Viscous Force ($F_v$​): Upward force due to the viscosity of glycerin. This balances the remaining downward force at terminal velocity.

From equilibrium at terminal velocity:

$$F_v = F_g - F_b$$​

The density of glycerin ($\rho_{\text{glycerin}}$) is half the density of the sphere ($\rho_{\text{sphere}}$):

$$\rho_{\text{glycerin}} = \frac{\rho_{\text{sphere}}}{2}$$​​

The mass $M$ of the sphere is related to its density and volume:

$$M = \rho_{\text{sphere}} V$$

Substitute $\rho_{\text{glycerin}} = \frac{\rho_{\text{sphere}}}{2}$​​ into $F_b$:

$$F_b = \frac{\rho_{\text{sphere}}}{2} V g = \frac{M}{2} g$$

Using $F_v = F_g - F_b$:

$$F_v = Mg - \frac{M}{2}g$$

Simplify:

$$F_v = \frac{M}{2} g$$

The viscous force acting on the sphere is:

$$\boxed{\frac{Mg}{2}}$$​​

When two droplets, each with a radius \(R\), merge to form a larger droplet, the total volume is conserved. Therefore:

\(2 \times \frac{4}{3} \pi R^3=\frac{4}{3} \pi r^3\)

Solving for \(r\), the radius of the larger drop, we find:

\(r=2^{1 / 3} R\)

The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension \(T\) :

\(U_i=2 \times 4 \pi R^2 T\)

The surface energy of the larger droplet is given by:

\(U_f=4 \pi r^2 T=4 \pi R^2 T\left(2^{2 / 3}\right)\)

The energy released in the process is the difference between the initial and final surface energy:
Heat lost \(=U_i-U_f=4 \pi R^2 T\left[2-2^{2 / 3}\right]\)

The pressure difference inside an air bubble submerged in water is due to two contributions:

1. Due to surface tension: $\Delta P_{\text{surface tension}} = \frac{2T}{r}$​, where $T$ is the surface tension and $r$ is the radius of the bubble.
2. Due to water column: $\Delta P_{\text{water}} = \rho g h$, where $\rho$ is the density of water, $g$ is gravitational acceleration, and $h$ is the depth of the bubble below the surface.

$$\Delta P_{\text{surface tension}} = \frac{2T}{r}$$​

Given:

$$T = 70 \times 10^{-3} \, \text{N/m}, \quad r = 2 \, \text{cm} = 0.02 \, \text{m}$$,

$$\Delta P_{\text{surface tension}} = \frac{2 \times 70 \times 10^{-3}}{0.02}$$,

​ $$\Delta P_{\text{surface tension}} = 7 \, \text{N/m}^2$$

$$\Delta P_{\text{water}} = \rho g h$$

Given:

$$\rho = 1000 \, \text{kg/m}^3, \quad g = 9.8 \, \text{m/s}^2, \quad h = 20 \, \text{cm} = 0.2 \, \text{m}$$

$$\Delta P_{\text{water}} = 1000 \times 9.8 \times 0.2$$

$\mathrm{\Delta }{P}_{\text{water}}=1960{\text{N/m}}^2$

The total pressure difference is the sum of the two contributions:

$$\Delta P = \Delta P_{\text{surface tension}} + \Delta P_{\text{water}}$$

$\mathrm{\Delta }P=7+1960=1967{\text{N/m}}^2$

Convert to standard form:

$\mathrm{\Delta }P=0.01967\times 10^5{\text{N/m}}^2\text{≈}0.02\times 10^5{\text{N/m}}^2$

$\boxed{\left(b\right)0.02\times 10^5{\text{N/m}}^2}$​

$\sigma =$ Density of ball

$$\begin{gathered} V_T=\frac{2R^2}{2\eta }(\sigma -\rho )g \\ {V}_T\propto R^2 \end{gathered}$$

As temperature changes Co-efficient of viscosity changes.

By continuity equation:

$A_1{v}_1=A_2{v}_2$

$A_2=30{ \mathrm{cm}}^2,v_1=50 \mathrm{cm}/\mathrm{s}$

$A_1=10\times 15{ \mathrm{mm}}^2=150{ \mathrm{mm}}^2=1.5{ \mathrm{cm}}^2$

$v_2=\frac{A_1{v}_1}{A_2}=\frac{30\times 50}{1.5}=1000 \mathrm{cm}/\mathrm{s}$

$\therefore v_2=10 \mathrm{m}/\mathrm{s}$

Let's convert the given values to decimeters ($\text{dm}$) for easier calculation since the volume is in ${\text{dm}}^3$:

Radius, $r=40\text{ cm}=4\text{ dm}$

Volume, $V=528{\text{ dm}}^3$

Using the formula for the volume of a cylinder ($V=\pi r^{2}H$):

So, the height of the vessel is $105\text{ cm}$.

Determine the height of the hole from the ground (\(y\))

The vessel is placed on a block of the same height ($105\text{ cm}$).

Total height of the top of the water level from the ground = $105\text{ cm}+105\text{ cm}=210\text{ cm}$.

The hole is $70\text{ cm}$ below the top ($h=70\text{ cm}$).

Height of the hole from the ground, $y=210\text{ cm}-70\text{ cm}=140\text{ cm}$.

Calculate the horizontal range (\(R\))

The formula for the horizontal range of efflux is:

Correct Option: B ($140\sqrt{2}$)

Statement 1: False, as the surface tension of soap is less than water.

Statement 2: True

The surface tension of soap solution is less than that of water because soap reduces the cohesive forces between water molecules. However, hot water flows faster than cold water due to its lower viscosity, which decreases with an increase in temperature.

$W=S\times 4\pi R^2\times \left(n^{1/3}-1\right)$
where n is the number of small drops.
For 27 drops:

$$\begin{gathered} \text{ Since }{27}^{1/3}=3, \\ 10=S4\pi R^2(3-1)=S4\pi R^2\times 2 \end{gathered}$$
For 64 Drops:
Similarly,

$$\begin{gathered} W^{\text{'}}=S4\pi R^2\left({64}^{1/3}-1\right) \\ \text{ Therefore, } \\ \frac{W^{\text{'}}}{W}=\frac{4-1}{3-1}=\frac{3}{2} \\ W^{\text{'}}=10\times \frac{3}{2}=15J \end{gathered}$$

$$\begin{gathered} \text{ Apply Bernoulli's eqn } \\ \frac{P_1}{\rho g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g} \\ \frac{P_1}{\rho g}+0=\frac{P_2}{\rho g}+\frac{V^2}{2g} \\ V=\sqrt{\frac{2\left(P_1-P_2\right)}{\rho }} \\ V\propto \sqrt{\left(P_1-P_2\right)} \end{gathered}$$

$$\begin{gathered} v_T=\frac{2}{9}\frac{\left({\rho }_0-{\rho }_l\right)r^{2}g}{\eta } \\ \eta =\frac{2}{9}\left(\frac{7825-925}{2.45\times {10}^{-2}}\right)\times {\left(1.8\right)}^2\times {10}^{-6}\times 9.8 \\ \eta \approx 1.99 \end{gathered}$$

At terminal velocity, forces balance: Weight D = D Buoyant force + Viscous force

$Mg=F_{\text{buoyancy }}+F_{\text{viscous }}$

Buoyant force: Since glycerine's density is half of the ball's density:

$F_{\text{buayancy }}=\frac{1}{2}Mg$

Viscous force:

$F_{\text{visceus }}=Mg-\frac{1}{2}Mg=\frac{Mg}{2}$

Pressure in the air bubble:

$P_{inside}=P_0+\rho gh+\frac{2T}{R}$

Given:

$P_{inside}-P_0=2100$

Hydrostatic pressure:

$$\begin{gathered} P_{\text{inside }}-P_0=\rho gh+\frac{2T}{R} \\ 2100=1000\times 10\times 0.2+\frac{2T}{0.001} \\ 2100=2000+\frac{2T}{0.001} \\ T=\frac{1}{20}=0.05 \end{gathered}$$

Thus, the correct answer is 0.05 .

Initial radius:

$R=1mm={10}^{-3}m$

Volume conservation:

$\frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3$

$r=\frac{R}{8}$Initial surface area: $A_i=4\pi R^2$
Final surface area:

$A_f=512\times 4\pi r^2=512\times 4\pi {\left(\frac{R}{8}\right)}^2$

$=8\times 4\pi R^2=32\pi R^2$

Increase: $\Delta A=A_f-A_i=(32-4)\pi R^2=28\pi R^2$

Change in surface energy:

$\Delta U=T\Delta A=0.08\times 28\pi {\left({10}^{-3}\right)}^2$

$\approx 7.04\times {10}^{-6}J$

$\alpha \approx 7$

At terminal velocity, forces balance: Weight D = D Buoyant force + Viscous force

$Mg=F_{\text{buoyancy }}+F_{\text{viscous }}$

Buoyant force: Since glycerine's density is half of the ball's density:

$F_{\text{buayancy }}=\frac{1}{2}Mg$

Viscous force:

$F_{\text{visceus }}=Mg-\frac{1}{2}Mg=\frac{Mg}{2}$

This question is based on Bernoulli's equation, which states:

Since the pipe is horizontal ($h$h is constant), the equation simplifies to:

For points A and B:

Rearranging:

Divide through by $\rho/2$:

The correct option is:

Statement 1: False, as the surface tension of soap is less than water.

Statement 2: True

There are two liquid-air surfaces in bubble so

$\Delta P=2\left(\frac{2S}{R}\right)=\frac{4S}{R}$

Volume of big drop = 1000 × Volume of small drop

$\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3$

$R=10r$

Surface Energy $=T\Delta A$

$\frac{U_{1000}}{U_B}=\frac{A_1}{A_2}=\frac{4\pi r^2\times 1000}{4\pi R^2}$

$\frac{U_{\text{small}}}{U_{\text{big}}}=\frac{r^2\times 1000}{100r^2}=\frac{10}{1}$

Statement I is correct as we know contact angle depends on cohesine and adhesive forces.
Statement II is incorrect because height of liquid is given by h $-=\frac{2T\cos 0_c}{\rho gr}$where r is radius of Tube (assuming length of capillary is sufficient)

Statement I is correct as we know contact angle depends on cohesine and adhesive forces.
Statement II is incorrect because height of liquid is given by h $-=\frac{2T\cos 0_c}{\rho gr}$where r is radius of Tube (assuming length of capillary is sufficient)

The work done is proportional to the change in surface area. The surface area of a sphere is $4 \pi r^2$.

1. Initial bubble area:
   The radius of the bubble is $R$. Its surface area is:

   $$A_{\text{initial}} = 4 \pi R^2$$

2. For 27 drops:

   • Each drop has radius $r_1$ such that the volume of the bubble equals the total volume of the drops: $$\frac{4}{3} \pi R^3 = 27 \cdot \frac{4}{3} \pi r_1^3 \implies r_1 = \frac{R}{3}$$​
   • Total surface area of 27 drops: $$A_{\text{27 drops}} = 27 \cdot 4 \pi r_1^2 = 27 \cdot 4 \pi \left( \frac{R}{3} \right)^2 = 12 \pi R^2$$
   • Change in surface area: $$\Delta A = A_{\text{27 drops}} - A_{\text{initial}} = 12 \pi R^2 - 4 \pi R^2 = 8 \pi R^2$$
   • Work done is proportional to $\Delta A$: $$W_1 = k \Delta A \implies k = \frac{10}{8 \pi R^2}$$​

3. For 64 drops:

   • Each drop has radius $r_2$​ such that: $$\frac{4}{3} \pi R^3 = 64 \cdot \frac{4}{3} \pi r_2^3 \implies r_2 = \frac{R}{4}$$​
   • Total surface area of 64 drops: $$A_{\text{64 drops}} = 64 \cdot 4 \pi r_2^2 = 64 \cdot 4 \pi \left( \frac{R}{4} \right)^2 = 16 \pi R^2$$
   • Change in surface area: $$\Delta A = A_{\text{64 drops}} - A_{\text{initial}} = 16 \pi R^2 - 4 \pi R^2 = 12 \pi R^2$$
   • Work done: $$W_2 = k \Delta A = \frac{10}{8 \pi R^2} \cdot 12 \pi R^2 = 15 \, \text{J}$$

Thus, the work done when splitting into 64 drops is 15 J.

Correct Answer: D) Statement I is false but Statement II is true
Short Explanation:
- Statement I is false: If the liquid neither rises nor falls in a capillary, it implies the contact angle is \(90^{\circ}\), not \(0^{\circ}\). A contact angle of \(0^{\circ}\) would mean the liquid wets the surface completely and rises in the capillary.
- Statement II is true: The contact angle depends on the intermolecular interactions between the specific solid and liquid materials, so it's indeed a property of both.

Correct Answer: D) Statement I is false but Statement II is true
Short Explanation:
- Statement I is false: If the liquid neither rises nor falls in a capillary, it implies the contact angle is \(90^{\circ}\), not \(0^{\circ}\). A contact angle of \(0^{\circ}\) would mean the liquid wets the surface completely and rises in the capillary.
- Statement II is true: The contact angle depends on the intermolecular interactions between the specific solid and liquid materials, so it's indeed a property of both.

The pressure difference inside an air bubble submerged in water is due to two contributions:

1. Due to surface tension: $\Delta P_{\text{surface tension}} = \frac{2T}{r}$​, where $T$ is the surface tension and $r$ is the radius of the bubble.
2. Due to water column: $\Delta P_{\text{water}} = \rho g h$, where $\rho$ is the density of water, $g$ is gravitational acceleration, and $h$ is the depth of the bubble below the surface.

$$\Delta P_{\text{surface tension}} = \frac{2T}{r}$$​

Given:

$$T = 70 \times 10^{-3} \, \text{N/m}, \quad r = 2 \, \text{cm} = 0.02 \, \text{m}$$,

$$\Delta P_{\text{surface tension}} = \frac{2 \times 70 \times 10^{-3}}{0.02}$$,

​ $$\Delta P_{\text{surface tension}} = 7 \, \text{N/m}^2$$

$$\Delta P_{\text{water}} = \rho g h$$

Given:

$$\rho = 1000 \, \text{kg/m}^3, \quad g = 9.8 \, \text{m/s}^2, \quad h = 20 \, \text{cm} = 0.2 \, \text{m}$$

$$\Delta P_{\text{water}} = 1000 \times 9.8 \times 0.2$$

$\mathrm{\Delta }{P}_{\text{water}}=1960{\text{N/m}}^2$

The total pressure difference is the sum of the two contributions:

$$\Delta P = \Delta P_{\text{surface tension}} + \Delta P_{\text{water}}$$

$\mathrm{\Delta }P=7+1960=1967{\text{N/m}}^2$

Convert to standard form:

$\mathrm{\Delta }P=0.01967\times 10^5{\text{N/m}}^2\text{≈}0.02\times 10^5{\text{N/m}}^2$

$\boxed{\left(b\right)0.02\times 10^5{\text{N/m}}^2}$​

The ratio k is defined in terms of the angles at two surfaces A and B.
$k=\frac{\cos {\theta }_A}{\cos {\theta }_B}$
If k is negative, one surface is convex and the other surface is concave.
If k is zero, one surface is flat while the other surface is convex or concave.

Surface tension is property of surface molecules not interior molecules. In case of liquid as temperature increases surface tension decreases. In case of gases due to increase in temperature collision will be more frequent hence viscosity increases.

$W=S\times 4\pi R^2\times \left(n^{1/3}-1\right)$
where n is the number of small drops.
For 27 drops:

$$\begin{gathered} \text{ Since }{27}^{1/3}=3, \\ 10=S4\pi R^2(3-1)=S4\pi R^2\times 2 \end{gathered}$$
For 64 Drops:
Similarly,

$$\begin{gathered} W^{\text{'}}=S4\pi R^2\left({64}^{1/3}-1\right) \\ \text{ Therefore, } \\ \frac{W^{\text{'}}}{W}=\frac{4-1}{3-1}=\frac{3}{2} \\ W^{\text{'}}=10\times \frac{3}{2}=15J \end{gathered}$$

The surface tension of soap solution is less than that of water because soap reduces the cohesive forces between water molecules. However, hot water flows faster than cold water due to its lower viscosity, which decreases with an increase in temperature.