Mechanical Properties of Fluids

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.

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

Statement 2: True

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

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

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:

$$\boxed{\frac{Mg}{2}}$$​​

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 .

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

There are two liquid-air surfaces in bubble so

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

$W=S\times 4\pi R^2\times \left(n^{1/3}-1\right)$
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,

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

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

Statement 2: True

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.

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 .

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:

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:

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

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.

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

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,

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

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

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:

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