Fluid Mechanics

Length of side of cube

$ =40 \mathrm{~cm}=0.4 \mathrm{~m} $

When only cube is floating, $\frac{1}{4}$ th of its volume is submerged in water.

$ \begin{aligned} \therefore \quad m_c g & =\rho g \times\left[\frac{1}{4} r\right] \\ m_c & =\frac{1}{4} \rho V=\frac{1}{4} \times 1000 \times(0.4)^3 \\ m_c & =16 \mathrm{~kg} \end{aligned} $

When disc is placed on the cube, $\frac{2}{5}$ th of cube is immersed in water.

$ \begin{aligned} \therefore \quad & {\left[m_c+m_d \mathbb{g}=\rho g\left[\frac{2}{5} V\right]\right.} \\ & m_d=\frac{2}{5} \times 1000 \times(0.4)^3-16 \\ & m_d=25.6-16=9.6 \mathrm{~kg} \end{aligned} $

The maximum length (height) to which a liquid rises in a capillary tube

$ \begin{aligned} h & =\frac{2 T \cos \theta}{\rho g R}=\frac{2 \times 0.07 \times \cos 0^{\circ}}{10^3 \times 10 \times 0.5 \times 10^{-3}} \\ & =0.028 \mathrm{~m}=2.8 \mathrm{~cm} \end{aligned} $

As block is floating

⇒ Weight of block $=$ Weight of water displaced.

$ \begin{array}{ll} \Rightarrow & (1-V) \frac{3}{4} \times \rho_w \times g=\frac{1}{2} \times \rho_w \times g \\ \Rightarrow & 1-V=\frac{4}{6} \end{array} $

$ \begin{aligned} \Rightarrow \quad V & =1-\frac{4}{6}=\frac{2}{6}=\frac{1}{3} \mathrm{Ltr} \\ & =\frac{1000}{3} \mathrm{~mL} \\ & =333.33 \mathrm{~mL} \end{aligned} $

Step 1: Find the acceleration

We know the final speed $v = 18~\mathrm{m/s}$, the starting speed $u = 0$, and the height $h = 20~\mathrm{m}$.

Use the formula: $v^2 = u^2 + 2 a h$

Put the values in: $18^2 = 0^2 + 2 a \times 20$

This gives: $a = \frac{18 \times 18}{2 \times 20} = 8.1~\mathrm{m/s}^2$

Step 2: Find the time taken

Use the formula: $v = u + a t$

Plug in the numbers: $18 = 0 + 8.1 \times t$

Solve for $t$:

$t = \frac{18}{8.1} = 2.22~\mathrm{s} \approx 2.2~\mathrm{s}$

When you pull the wires apart, the water film stretches, making its area bigger.

A thin film has two surfaces (top and bottom), so the increase in area is:

$ \Delta A = 2 \times L \times \Delta d $

Let's define what we have:

Length of each wire, $ L = 8\,\mathrm{cm} = 0.08\,\mathrm{m} $

Starting distance, $ d_1 = 0.6\,\mathrm{cm} = 0.006\,\mathrm{m} $

Final distance, $ d_2 = 0.8\,\mathrm{cm} = 0.008\,\mathrm{m} $

Change in distance:

$ \Delta d = d_2 - d_1 = 0.008 - 0.006 = 0.002\,\mathrm{m} $

Calculate the change in area:

$ \Delta A = 2 \times 0.08 \times 0.002 = 3.2 \times 10^{-4}\,\mathrm{m}^2 $

The work done (energy needed) is given by:

$ W = T \Delta A $

Here, surface tension $ T = 0.07\,\mathrm{N\,m}^{-1} $

So,

$ W = 0.07 \times 3.2 \times 10^{-4} = 0.224 \times 10^{-4}\,\mathrm{J} $

$ W = 22.4 \times 10^{-6}\,\mathrm{J} = 22.4 \,\mu \mathrm{J} $

$ \begin{aligned} &\text { Viscous force }\\ &\begin{aligned} F & =6 \pi \eta r v \\ & =6 \times 3.14 \times 2 \times 10^{-5} \times 0.5 \times 10^{-3} \times 0.7 \\ & =13.2 \times 10^{-8} \mathrm{~N} \end{aligned} \end{aligned} $

Step 1: Find Surface Area Relationship

The bubble’s surface area at the top is $125\%$ more than at the bottom. This means:

$ A_{\text {top }} = A_{\text {bottom }} + 1.25A_{\text {bottom }} = 2.25A_{\text {bottom }} $

Step 2: Connect Surface Area and Volume

The volume ($V$) of a bubble is related to its surface area ($A$) by: $ V \propto A^{3/2} $

Step 3: Express Volume Change

Let $V_1$ be the volume at the bottom, and $V_2$ at the top. Then,

$ \frac{V_2}{V_1} = \left(\frac{A_2}{A_1}\right)^{3/2} = (2.25)^{3/2} = 3.375 $

Step 4: Use Boyle’s Law for Pressure and Volume

Since the water temperature is the same, Boyle’s law applies: $ p_1 V_1 = p_2 V_2 $ $ \frac{p_1}{p_2} = \frac{V_2}{V_1} = 3.375 $

Step 5: Write Out Pressure at Bottom and Top

At the bottom, pressure is $ p_1 = \text{atmospheric pressure} + \text{pressure from water column with height } h $ Atmospheric pressure is equal to water pressure from $10\,\mathrm{m}$ column. So: $ p_1 = 10 + h $ At the top, the pressure is just the atmospheric pressure: $ p_2 = 10 $

Step 6: Set Up the Equation and Solve for $h$

So, $ \frac{p_1}{p_2} = \frac{10 + h}{10} $ Set equal to $3.375$: $ \frac{10 + h}{10} = 3.375 $ $ 10 + h = 33.75 $ $ h = 23.75\,\mathrm{m} $

To find the work done to make a soap bubble bigger, we use the formula:

$ W = 2T \cdot \Delta A $

Step 1: Find Change in Surface Area for $W_1$

The surface area of a sphere is $4\pi r^2$. When the radius goes from $r$ to $2r$, the change in area is:

$ \Delta A = 4\pi(2r)^2 - 4\pi r^2 $

This simplifies to:

$ = 16\pi r^2 - 4\pi r^2 = 12\pi r^2 $

Step 2: Find Work Done $W_1$

Plug the change in area into the work formula:

$ W_1 = 2T \cdot 12\pi r^2 = 24\pi T r^2 $

Step 3: Find Change in Surface Area for $W_2$

Now, change the radius from $2r$ to $3r$:

$ \Delta A = 4\pi(3r)^2 - 4\pi(2r)^2 $

$ = 36\pi r^2 - 16\pi r^2 = 20\pi r^2 $

Step 4: Find Work Done $W_2$

Plug in this change in area:

$ W_2 = 2T \cdot 20\pi r^2 = 40\pi T r^2 $

Step 5: Ratio $W_1 : W_2$

$ W_1 : W_2 = 24 : 40 = 3 : 5 $

Height of water $h=20 \mathrm{~cm}$

$ =20 \times \frac{1}{100} \mathrm{~m}=0.2 \mathrm{~m} $

Velocity of efflux

$ \begin{aligned} & v=\sqrt{2 g h}=\sqrt{2 \times 10 \mathrm{~m} / \mathrm{s}^2 \times 0.2 \mathrm{~m}} \\ & v=\sqrt{4 \mathrm{~m}^2 / \mathrm{s}^2}=2 \mathrm{~ms}^{-1} \end{aligned} $

Volume flow rate

$ \begin{aligned} & Q=A \times V \\ & Q=1 \mathrm{~m}^2 \times 2 \mathrm{~m} / \mathrm{s}=2 \mathrm{~m}^3 / \mathrm{s} \end{aligned} $

Volume of water $V=Q \times t$

$ V=2 \mathrm{~m}^3 / \mathrm{s} \times 0.6 \mathrm{~s}=1.2 \mathrm{~m}^3 $

Mass of the water

$ \begin{aligned} m & =\rho \times V \\ m & =1000 \mathrm{~kg} / \mathrm{m}^3 \times 1.2 \mathrm{~m}^3 \\ & =1200 \mathrm{~kg}=1.2 \mathrm{~g} \end{aligned} $

For stream line flow

$ R_e<1000 $

Thus, for $R_e=900$, flow of liquid is streamlined.

$ \begin{aligned} & d=3 \mathrm{~cm} \\ & r=\frac{3}{2} \times 10^{-2} \mathrm{~m} \\ & T=0.035 \mathrm{Nm}^{-1} \end{aligned} $

For soap bubble, both inner and outer surface contributes, so increase in surface area is

$ \Delta A=2 \times 4 \pi r^2=8 \pi r^2 $

Work done $=$ Surface tension $\times$ Increase in area

$ \begin{aligned} W & =T \times \Delta A \\ & =0.035 \times 8 \times \pi \times\left(\frac{3}{2} \times 10^{-2}\right)^2 \\ & \approx 1.98 \times 10^{-4} \mathrm{~J} \approx 198 \mu \mathrm{~J} \end{aligned} $

We know that terminal velocity

$ \begin{array}{ll} & v=\frac{2 r^2(\rho-\sigma) g}{9 \eta} \\ \Rightarrow & v \propto r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

But mass of sphere

$ \begin{array}{ll} & m=\frac{4}{3} \pi r^3 \rho \\ \Rightarrow & r^3 \propto m \\ \Rightarrow & r \propto m^{1 / 3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & v \propto\left(m^{1 / 3}\right)^2 \\ \Rightarrow \quad & v \propto m^{2 / 3} \\ \Rightarrow \quad & \frac{v_2}{v_1}=\left(\frac{m_2}{m_1}\right)^{2 / 3} \end{aligned} $

$ \begin{aligned} & =\left(\frac{64}{8}\right)^{2 / 3}=(8)^{2 / 3} \\ & =\left(2^3\right)^{2 / 3}=2^2=4 \\ \Rightarrow \quad v_2 & =4 v_1=4 \times 3=12 \mathrm{~cm} / \mathrm{s} \end{aligned} $

Given:

• Length of the wire, $ l = 20~\mathrm{cm} = 0.20~\mathrm{m} $

• Force required to keep the wire in equilibrium, $ F = 1.456 \times 10^{-2}~\mathrm{N} $

We are asked to find the surface tension $ T $ of water.

Concept:

When a wire is in contact with a liquid surface, two surfaces are stretched — one on each side of the wire (front and back).

The total upward force due to surface tension is:

$ F = 2 T l $

(where the factor 2 accounts for the two surfaces).

Calculation:

$ T = \dfrac{F}{2 l} $

Substitute the values:

$ T = \dfrac{1.456 \times 10^{-2}}{2 \times 0.20} $

$ T = \dfrac{1.456 \times 10^{-2}}{0.40} $

$ T = 3.64 \times 10^{-2}~\mathrm{N/m} $

✅ Final Answer:

$ \boxed{T = 0.0364~\mathrm{N/m}} $

Correct Option: (B) $0.0364~\mathrm{Nm^{-1}}$

When two soap bubbles of radii $r_1$ and $r_2$ coalesce isothermally in vacuum, the radius $R$ of the new bubble is given as

$ R=\sqrt{r_1^2+r_1^2}=\sqrt{2^2+2^2}=2 \sqrt{2} \mathrm{~cm} $

The correct answer is:

✅ Option B: gases increase but liquids decrease

Explanation:

• For liquids, viscosity is mainly due to intermolecular attraction. As temperature increases, these attractive forces weaken, and molecules move more freely — → viscosity decreases.

• For gases, viscosity arises from molecular momentum transfer. As temperature increases, molecular speed increases, enhancing momentum exchange — → viscosity increases.

As temperature is constant, for air in bubble,

$ p_1 V_1=p_2 V_2 $

Given,

$ \begin{aligned} & p_1=10 \times \rho_w \times g \\ & p_2=(10+7.28) \rho_w \times g \\ & V_1=\frac{4}{3} \pi r_1^3 \text { and } V_2=\frac{4}{3} \pi r_2^3 \end{aligned} $

Now by $p_1 V_1=p_2 V_2$

$ \begin{aligned} \Rightarrow \quad \frac{V_2}{V_1} & =\left(\frac{r_2}{r_1}\right)^3=\frac{p_1}{p_2} \\ \Rightarrow \quad \frac{r_2}{r_1} & =\left(\frac{p_1}{p_2}\right)^{\frac{1}{3}}=\left(\frac{10}{17.28}\right)^{\frac{1}{3}} \\ & =\left(\frac{1000}{1728}\right)^{\frac{1}{3}}=\left(\frac{125}{216}\right)^{\frac{1}{3}}=\frac{5}{6} \end{aligned} $

Step 1: Find the volume of water hitting the wall each second.

The formula for flow rate (volume per second) is:

$ \text{Volume per second} = \text{Area} \times \text{Velocity} $

Here, Area = $2 \times 10^{-3}~\mathrm{m}^2$ and Velocity = $12~\mathrm{ms}^{-1}$

$ \text{Volume per second} = 2 \times 10^{-3} \times 12 = 24 \times 10^{-3}~\mathrm{m}^3/\mathrm{s} $

Step 2: Find the mass of water hitting the wall each second.

To get the mass per second, multiply the volume flow by the density of water ($\rho = 1000~\mathrm{kg/m^3}$):

$ \text{Mass per second} = \text{Density} \times \text{Volume per second} $

$ = 1000 \times 24 \times 10^{-3} = 24~\mathrm{kg/s} $

Step 3: Calculate the force on the wall.

Force is the change in momentum per second. The water hits the wall at velocity $12~\mathrm{ms}^{-1}$ and stops instantly.

$ \text{Force} = \text{Mass per second} \times \text{Velocity} $

$ = 24 \times 12 = 288~\mathrm{N} $

Step 1: Find Extra Pressure Inside a Bubble

The extra (excess) pressure inside a soap bubble is $p = \frac{4T}{r}$, where $T$ is surface tension and $r$ is the radius.

So for bubble $A$: $p_A = \frac{4T}{r_1}$

And for bubble $B$: $p_B = \frac{4T}{r_2}$

Step 2: Use the Gas Law for Bubbles

The usual gas equation is $pV = nRT$. Here, $p$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature.

Since the temperature is constant, $pV \propto n$. The number of moles $n$ is related to the mass $m$ of the gas. So, $pV \propto m$.

Step 3: Work Out the Mass for Each Bubble

For bubble $A$:

The volume of a sphere is $V_A = \frac{4}{3}\pi r_1^3$.

So $p_A V_A \propto m_A$ becomes:

$\frac{4T}{r_1} \times \frac{4}{3}\pi r_1^3 \propto m_A$

This simplifies to $m_A \propto r_1^2$.

For bubble $B$:

Similarly, $V_B = \frac{4}{3}\pi r_2^3$.

So $\frac{4T}{r_2} \times \frac{4}{3}\pi r_2^3 \propto m_B$

This gives us $m_B \propto r_2^2$.

Step 4: Find the Ratio of the Masses

The ratio is $\frac{m_A}{m_B} = \frac{r_1^2}{r_2^2}$.

$ \text { } \begin{aligned} \frac{4}{3} \pi R^3 & =10^6 \times \frac{4}{3} \pi r^3 \\ R & =100 r \\ \Rightarrow \quad r & =\frac{R}{100} \end{aligned} $

$ \begin{aligned} &\therefore \text { Change in surface energy }\\ &\begin{aligned} & \Delta E=T \times \Delta A \\ & =35 \times 10^{-3}\left[10^6 \times 4 \pi r^2-4 \pi R^2\right] \\ & =35 \times 10^{-3}\left[10^6 \times 4 \pi \times\left(\frac{R}{100}\right)^2-4 \pi R^2\right] \\ & =35 \times 10^{-3}\left[400 \pi R^2-4 \pi R^2\right] \\ & =35 \times 10^{-3} \times 396 \pi R^2 \\ & =35 \times 10^{-3} \times 396 \times \frac{22}{7} \times 1 \times 10^{-4} \\ & =4356 \times 10^{-6} \mathrm{~J}=4356 \mu \mathrm{~J} \end{aligned} \end{aligned} $

Given:

• Mass of the aeroplane, $ m = 4.5 \times 10^4 \, \mathrm{kg} $

• Total wing area, $ A = 600 \, \mathrm{m^2} $

• Acceleration due to gravity, $ g = 10 \, \mathrm{m/s^2} $

The aeroplane is travelling at a constant height, meaning lift = weight.

Step 1. Lift = Pressure difference × Area

$ L = \Delta P \times A $

Step 2. Lift = Weight

$ L = m g $

So,

$ \Delta P \times A = m g $

Step 3. Solve for pressure difference

$ \Delta P = \frac{m g}{A} $

Substitute the known values:

$ \Delta P = \frac{4.5 \times 10^4 \times 10}{600} $

$ \Delta P = \frac{4.5 \times 10^5}{600} $

$ \Delta P = 750 \, \mathrm{N/m^2} $

✅ Answer: $\boxed{750 \, \mathrm{N/m^2}}$

Correct Option: D

Since, pressure is same on the both side, thus

$ \begin{aligned} \frac{F_1}{A_1} & =\frac{F_2}{A_2} \\ \Rightarrow \quad F_2 & =\frac{F_1 A_2}{A_1}=\frac{250 \times \pi r_2^2}{\pi r_1^2} \\ & =250 \times\left(\frac{r_2}{r_1}\right)^2=250 \times\left(\frac{50}{5}\right)^2 \\ & =25000 \mathrm{~N}=25 \mathrm{kN} \end{aligned} $

If $d$ be the diameter of small droplets, then

$ \begin{aligned} \frac{4}{3} \pi\left(\frac{D}{2}\right)^3 & =\frac{4}{3} \pi\left(\frac{d}{2}\right)^3 \times 3375 \\ D & =15 d \end{aligned} $

$\therefore$ Change in surface energy

$ \begin{aligned} \Delta W & =S \times \Delta A \\ & =S \times\left[3375 \times 4 \pi\left(\frac{d}{2}\right)^2-4 \pi\left(\frac{D}{2}\right)^2\right] \\ & =S \times\left[3375 \times 4 \pi\left(\frac{D}{15 \times 2}\right)^2-\pi D^2\right] \\ & =S\left(15 D^2 \pi-\pi D^2\right) \\ & =14 \pi D^2 S \end{aligned} $

Reynolds number

$ \begin{aligned} R_e & =\frac{\rho v d}{\eta} \\ & =\frac{10^3 \times 0.2 \times 0.04}{10^{-3}}=8000 \end{aligned} $

Since, $R_z>2000$

Thus, flow of water is turbulent.

Given,

$ \frac{p_{1}}{p_{2}}=\frac{2}{3} $

We know that, the excess pressure inside a soap bubble is given by the formula

$ \Delta p=\frac{4 Y}{r} $

where, $ \Delta p= $ excess pressure, $ Y $ is the surface tension of the soap and $ r $ is the radius of the bubble.

Using Eq. (i), we can write

$ \frac{\frac{4 Y}{r_{1}}}{\frac{4 Y}{r_{2}}}=\frac{2}{3} \Rightarrow \frac{r_{2}}{r_{1}}=\frac{2}{3} $

The ratio of the volumes of the soap bubble is

$ \begin{array}{l} \frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}} \\\\ \frac{V_{1}}{V}=\left(\frac{r_{1}}{r_{2}}\right)^{3} \end{array} $

Using Eq. (ii), we get

$ \begin{aligned} \frac{V_{1}}{V_{2}} & =\left(\frac{3}{2}\right)^{3} \\\\ V_{1}: V_{2} & =27: 8 \end{aligned} $

According to Bernouli's principle, the pressure difference between the top and bottom surfaces of the wing can be calculated as

$ p_{\text {bottom }}-p_{\text {top }}=\frac{1}{2} \rho\left(v_{\text {top }}^{2}-v_{\text {bottom }}^{2}\right) $

Given, $ \rho=1.3 \mathrm{~kg} / \mathrm{m}^{3}, v_{\text {top }}=50 \mathrm{~m} / \mathrm{s} $,

$ v_{\text {bottom }}=40 \mathrm{~m} / \mathrm{s} $

After putting values,

$ \begin{aligned} p_{\text {bottom }}-p_{\text {top }} & =\frac{1}{2} \times 1.3 \times\left[(50)^{2}-(40)^{2}\right] \\\\ & =\frac{1}{2} \times 1.3 \times 900 \\\\ & =585 \mathrm{~N} / \mathrm{m}^{2} \end{aligned} $

The lift force is the pressure difference multiplied by the area of the wing.

$ \begin{aligned} F_{\text {lift }} & =\left(p_{\text {bottom }}-p_{\text {top }}\right) \cdot A \\\\ & =585 \times 10=5850 \mathrm{~N} \end{aligned} $

The weight of the aeroplane,

$ w=m g=500 \times 9.8=4900 \mathrm{~N} $

$ \therefore $ Lift force is greater than the weight.

So, the net upward force is,

$ F_{\text {net }}=F_{\text {lift }}-w=5850-4900=950 \mathrm{~N} $

Since the net force acting on the aeroplane is upward, the aeroplane will gain altitude.

$ \begin{aligned} & \rho=\text { density } \\\\ & \rho_A>\rho_B>\rho_C \end{aligned} $

We know the pressure and density is related as $ p=\rho g h $

$ p_{A}=\rho_{A} g h, p_{B}=\rho_{B} g h, p_{C}=\rho_{C} g h $

As $ g $ and $ h $ are constant.

$ \begin{array}{ll} \therefore & p_{A} \propto \rho_{A} p_{B} \propto \rho_{B}, p_{C} \propto \rho_{C} \\\\ \therefore & p_{A} > p_{B} > p_{C} \end{array} $

Hence, vessel $ A $ has maximum pressure.

Given,

Oil density $ =0.9 \mathrm{~g} / \mathrm{cm}^{3} $

Water density $ =1 \mathrm{~g} / \mathrm{cm}^{3} $

Weight of block = weight of oil displaced + weight of water displacd

Let $ m $ be mass of block

$ m g= $ (Volume of water displace $ \times $ Density of water $ \times g)+( $ Volume of oil displace $ \times $ Density of oil $ \times g $ )

$ \begin{array}{l} m=7 \times 10 \times 10 \times 0.9+3 \times 10 \times 10 \times 1 \\\\ m=630+300 \\\\ m=930 \mathrm{~g} \end{array} $

Given, radius of bubble, $r=0.5 \mathrm{~cm}$

$ =0.5 \times 10^{-2} \mathrm{~m} $

Height of air column, $h=4 \mathrm{~mm}$

$ =4 \times 10^{-3} \mathrm{~m} $

density of oil, $\rho=900 \mathrm{~kg} \mathrm{~m}^{-3}$

According to question;

Excess pressure in bubble

$=$ Pressure due to air column

$ \begin{aligned} \frac{4 S}{R} & =\rho g h \\\\ \Rightarrow \quad S & =\frac{\rho g h R}{4} \\\\ & =\frac{900 \times 10 \times 4 \times 10^{-3} \times 0.5 \times 10^{-2}}{4} \\\\ & =4.5 \times 10^{-2} \mathrm{Nm}^{-1} \end{aligned} $

Given, Water flow rate $=12 \pi \frac{\text { litre }}{\mathrm{min}}$

$ =\frac{12 \pi \times 10^{-3} \mathrm{~m}^3}{1 \times 60 \mathrm{~s}}=\frac{\pi}{5} \times 10^{-3} \mathrm{~cm}^3 / \mathrm{s} $

For one second; volume of water $=\pi r^2 h$

$ \begin{aligned} & \Rightarrow \quad \frac{\pi}{5} \times 10^{-3} \mathrm{~m}^3=\pi\left(\frac{d}{2}\right)^2 \times h \\\\ & \Rightarrow \quad \frac{10^{-3}}{5} \times \frac{1}{\left(1 \times 10^{-2}\right)^2}=h \quad(\because d=2 \mathrm{~cm}) \\\\ & \quad=\frac{10^{-3} \times 10^4}{5}=h \\\\ & \Rightarrow \quad h=2 \mathrm{~m} \end{aligned} $

$\text { or velocity }=\frac{h \mathrm{~m}}{1 \mathrm{~s}}=2 \mathrm{~m} / \mathrm{s} \mathrm{l}$

Area of hole $=2.4 \mathrm{~mm}^2$

Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ To determine water leaked from tank,

Using formula, $v=\sqrt{2 g h}$

$ \begin{aligned} & v=\sqrt{2 \times 10 \times 5} \\\\ & v=\sqrt{100}=10 \mathrm{~m} / \mathrm{s} \quad \ldots \text { (i) } \end{aligned} $

Now, we can determine flow water ( $Q$

$ \begin{aligned} Q & =v \times A \quad \ldots \text { (ii) } \\\\ A & =2.4 \mathrm{~mm}^2 \\\\ & =2.4 \times 10^{-6} \mathrm{~m}^2 \\\\ Q & =10 \times 2.4 \times 10^{-6} \\\\ & =2.4 \times 10^{-5} \mathrm{~m}^3 / \mathrm{s} \end{aligned} $

Volume $V$ leaked in 5 s ,

$ \begin{aligned} Q \times t & =2.4 \times 10^{-5} \times 5 \\\\ & =1.2 \times 10^{-4} \mathrm{~m}^3 \\\\ & =120 \times 10^{-6} \mathrm{~m}^3 \end{aligned} $

Hence, the volume of the water leaked in 5 s through the hole is $120 \times 10^{-6} \mathrm{~m}$.

Given:

Flow rate, $ Q = 7.5 \times 10^{-5} \, \text{m}^3/\text{s} $

Diameter of the tap, $ d = 1.5 \, \text{cm} = 0.015 \, \text{m} $

Coefficient of viscosity, $ \mu = 10^{-3} \, \text{Pa}\cdot\text{s} $

Reynolds Number Calculation

To determine the type of flow, we calculate the Reynolds number ($ R_e $) using the formula:

$ R_e = \frac{\rho v d}{\mu} $

where:

$\rho = 1000 \, \text{kg/m}^3$ (density of water),

$v$ is the velocity of the fluid,

$d$ is the diameter of the tap,

$\mu$ is the dynamic viscosity of the fluid.

Velocity of the Fluid

The velocity ($ v $) of the fluid can be calculated using the flow rate equation:

$ Q = Av $

where $ A = \pi \left(\frac{d}{2}\right)^2 $ is the cross-sectional area of the pipe.

$ v = \frac{Q}{A} = \frac{Q}{\pi \left(\frac{d}{2}\right)^2} $

Substituting the given values:

$ v = \frac{7.5 \times 10^{-5}}{\pi \times \left(\frac{0.015}{2}\right)^2} \approx 0.42 \, \text{m/s} $

Calculate Reynolds Number

Now, substitute the values into the Reynolds number formula:

$ R_e = \frac{1000 \times 0.42 \times 0.015}{10^{-3}} $

$ R_e = 6300 $

Interpretation of Reynolds Number

A Reynolds number less than 2000 indicates laminar flow.

A Reynolds number greater than 4000 indicates turbulent flow.

Since the calculated $ R_e = 6300 $, the flow is turbulent.

The pressure at the bottom of a vessel due to a liquid column is calculated using the formula:

$ p = \rho g h $

where:

$\rho$ is the density of the liquid,

$g$ is the acceleration due to gravity,

$h$ is the height of the liquid column.

In this scenario, both vessels $A$ and $B$ have the same height $h$ of water, and since the density of water and the acceleration due to gravity are constants, we can determine the pressure ratio at the bottom of the two vessels.

The ratio of pressures at the bottom of the vessels is:

$ \frac{p_A}{p_B} = \frac{\rho g h_A}{\rho g h_B} = \frac{\rho g h}{\rho g h} = 1 $

Thus, the ratio of pressures is:

$ p_A : p_B = 1 : 1 $

Given, pressure $=2000 \mathrm{~Pa}$

Let the pressure be $p$ at the second point and let velocity $v_2=v$

Applying equation of continuity,

$ \begin{aligned} A_1 v_1 & =A_2 v_2 \\ 10 \times 1 & =5 \times v \\ v & =2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Applying Bernoull's theorem at both point $A$ and $B$

$ \begin{gathered} p_1+\frac{1}{2} \rho v_1^2+\rho g h=p_2+\frac{1}{2} \rho v_2^2+\rho g h \\ \\ 2000+\frac{1}{2} \times 1000 \times 1^2 \\ = \\ p_2+\frac{1}{2} \times 1000 \times 4 \\ p_2= \end{gathered} $

Given:

Radius of ball, $ r $: $ 1 \times 10^{-4} \, \text{m} $

Density of the ball, $ \rho $: $ 10^4 \, \text{kg/m}^3 $

Density of water, $ \sigma $: $ 10^3 \, \text{kg/m}^3 $

Coefficient of viscosity of water, $ \eta $: $ 9.8 \times 10^{-6} \, \text{N s/m}^2 $

The terminal velocity $ v $ of a spherical object moving through a fluid can be calculated using the equation:

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

Substitute the given values into the equation:

$ v = \frac{2}{9} \times (1 \times 10^{-4})^2 \times \frac{(10^4 - 10^3)}{9.8 \times 10^{-6}} \times 9.8 $

Simplifying the expression gives:

$ v = \frac{2}{9} \times 10^{-8} \times \frac{9000}{9.8 \times 10^{-6}} \times 9.8 $

$ \Rightarrow v = 20 \, \text{m/s} $

The distance $ h $ required to reach this terminal velocity when falling in air is given by:

$ h = \frac{v^2}{2g} $

Substitute $ v = 20 \, \text{m/s} $ and $ g = 9.8 \, \text{m/s}^2 $:

$ h = \frac{(20)^2}{2 \times 9.8} $

$ \Rightarrow h = 20.4 \, \text{m} $

$ \text { According to Torricelli's theorem, } $

Case (i), Initial height, $h_1=h$

Final height, $h_2=\frac{h}{2}$

Case (ii), Initial height, $h_1{ }^{\prime}=\frac{h}{2}$

Final height, $h_2{ }^{\prime}=0$

From Eq. (i),

$ \begin{aligned} v & =\sqrt{2 g h} \\ \frac{d h}{d t} & =\sqrt{2 g h} \Rightarrow d t=\frac{d h}{\sqrt{2 g h}} \end{aligned} $

Integrating both side,

$ \begin{aligned} \int d t & =\int_0^h \frac{d h}{\sqrt{2 g h}} \\ t & =\sqrt{\frac{2 h}{g}} \end{aligned} $

For $t_1, t_1=\sqrt{\frac{2}{g}}\left(\sqrt{h}-\sqrt{\frac{h}{2}}\right)$

For $t_2, t_2=\sqrt{\frac{2}{g}}\left(\sqrt{\frac{h}{2}}-0\right)$

$ \begin{aligned} \text { Ratio, } \frac{t_1}{t_2} & =\frac{\sqrt{h}-\sqrt{\frac{h}{2}}}{\sqrt{\frac{h}{2}}} \\ \Rightarrow \quad \frac{t_1}{t_2} & =\frac{\sqrt{h}}{\sqrt{\frac{h}{2}}}-1 \\ \frac{t_1}{t_2} & =\sqrt{2}-1 \end{aligned} $

To find the fractional compression of water at the bottom of the river, let's work through the given data:

Given Information:

Depth of the river, $ h = 100 \, \text{m} $

Compressibility magnitude, $ \alpha = 0.5 \times 10^{-9} \, \text{N}^{-1} \text{m}^2 $

Acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $

Formula for Fractional Compression:

The fractional compression is given by the formula:

$ \left|\frac{\Delta V}{V}\right| = \frac{\Delta p}{B} $

where $ B $ is the bulk modulus of the water, and it's the reciprocal of compressibility:

$ B = \frac{1}{\alpha} $

Calculation Steps:

Fractional Compression Formula:

$ \left|\frac{\Delta V}{V}\right| = \Delta p \cdot \alpha $

Pressure Change ($\Delta p$):

The change in pressure due to the weight of the water column is:

$ \Delta p = \rho g h $

where:

$\rho$ (density of water) = $10^3 \, \text{kg/m}^3$

$g = 10 \, \text{m/s}^2$

$h = 100 \, \text{m}$

Calculate:

$ \Delta p = 10^3 \times 10 \times 100 = 10^6 \, \text{N/m}^2 $

Substitute into the Fractional Compression Formula:

$ \left|\frac{\Delta V}{V}\right| = 10^6 \times 0.5 \times 10^{-9} = 0.5 \times 10^{-3} $

Thus, the fractional compression of water at the bottom of the river is $0.5 \times 10^{-3}$.

To understand how the surface energy changes when two mercury drops merge, we start by considering the radii of the drops:

Let the radius of each smaller drop be $ r $.

When the two drops merge, the radius of the resulting bigger drop is $ R $.

The volume of a single spherical drop is given by $ \frac{4}{3} \pi r^3 $. Since the volumes combine when two droplets merge, we have:

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

Simplifying this equation, we find:

$ 2 \times r^3 = R^3 $

Taking the cube root of both sides, the radius of the larger drop becomes:

$ R = 2^{1/3} \cdot r $

Next, we compute the surface energy of the bigger drop. Surface energy ($ E $) is given by the product of the surface tension ($ T $) and the surface area ($ A $) of the drop:

$ E = T \cdot A $

The surface area of a sphere is $ 4 \pi R^2 $, hence for the larger drop:

$ A = 4 \pi R^2 = 4 \pi (2^{1/3} \cdot r)^2 = 4 \pi (2^{2/3} \cdot r^2) $

Solving further for $ E $:

$ E = T \cdot 4 \pi (2^{2/3} \cdot r^2) = 2^{8/3} \pi r^2 T $

Thus, the surface energy of the bigger drop is $ 2^{8/3} \pi r^2 T $.

To find the speed of ejection of the liquid from the holes, we begin with the given information:

Radius of the cylindrical tube, $ R = 2 \, \text{cm} = 0.02 \, \text{m} $

Number of fine holes, $ n = 50 $

Radius of each hole, $ r = 0.4 \, \text{mm} = 0.0004 \, \text{m} $

Speed of flow of liquid inside the tube, $ v_1 = 0.04 \, \text{m/s} $

First, calculate the cross-sectional area of the cylindrical tube:

$ A_1 = \pi R^2 = \pi (0.02)^2 = \pi (0.0004) \approx 0.00125 \, \text{m}^2 $

Next, calculate the total cross-sectional area of the 50 holes:

$ \begin{aligned} A_2 & = 50 \times \pi r^2 = 50 \times \pi (0.0004)^2 \\ & = 50 \times \pi (0.00000016) \\ & = 50 \times 0.000000502 \approx 0.0000251 \, \text{m}^2 \end{aligned} $

By using the conservation of mass (continuity equation) which states that the mass flow rate through the tube equals the mass flow rate through the holes, we have:

$ A_1 v_1 = A_2 v_2 $

Therefore, solve for $ v_2 $:

$ \begin{aligned} v_2 & = \frac{A_1 v_1}{A_2} \\ & = \frac{0.00125 \times 0.04}{0.0000251} \\ & = \frac{0.00005}{0.0000251} \approx 1.99 \approx 2 \, \text{m/s} \end{aligned} $

Thus, the speed of ejection of the liquid from the holes is approximately $ 2 \, \text{m/s} $.

$\begin{array}{lc} & S_1 \cos (\pi-\theta) l+S_2 l=S_3 l \\ \Rightarrow & -S_1 \cos \theta+S_2=S_3 \\ \Rightarrow & S_1 \cos \theta+S_3=S_2\end{array}$

To determine the energy released when 216 small liquid drops coalesce into a larger drop, follow these steps:

Given:

Number of small drops = 216

Surface area of each small drop = $A = 4 \pi r^2$

Surface tension of the liquid = $T$

Calculations:

Volumes:

Volume of one small drop: $\frac{4}{3} \pi r^3$

Total volume for 216 small drops: $216 \times \frac{4}{3} \pi r^3$

Volume of the large drop:

Since volume is conserved, the volume of the large drop is:

$ \frac{4}{3} \pi R^3 = 216 \times \frac{4}{3} \pi r^3 $

$ R^3 = 6r^3 \quad \Rightarrow \quad R = 6r $

Surface Areas:

Total surface area of 216 small drops:

$ 216 \times 4 \pi r^2 = 864 \pi r^2 $

Surface area of the large drop:

$ 4 \pi (6r)^2 = 144 \pi r^2 $

Decrease in Surface Area:

$ \Delta A = 864 \pi r^2 - 144 \pi r^2 = 720 \pi r^2 $

Energy Released:

Using the formula $E = T \cdot \Delta A$:

$ E = T \cdot 720 \pi r^2 $

$ E = 180 \cdot T \cdot 4 \pi r^2 $

$ E = 180 A T $

Thus, the energy released during the coalescence process is $180 A T$.

Given,

Atmospheric pressure height $=H$ Atmospheric pressure is given by $p_0=\rho g H$ where $\rho$ is density of water Let take the depth of water $=l$

So, pressure at depth $l$

$ \begin{aligned} & p=p_0+\rho g l \\ & p=\rho g(H+l) \end{aligned} $

Since, surrounding temperature is constant, process take place isothermally.

Using Boyle's Law

$ \begin{aligned} p_1 V_1 & =p_2 V_2\left[p_1=\rho g(H+l)\right. \\ p_2 & =\rho g H] \end{aligned} $

Also, $\quad V_2=5 V_1$

Put the value in Boyle's Law

$ V_1(\rho g(H)+\rho g l)=V_2(\rho g H) $

Solving this we get

$ l=4 H $

Hence, depth is $4 H$.

Given,

Surface area of each droplet $=10^{-7} \mathrm{~m}^2$

Surface tension $=0.07 \mathrm{~N} / \mathrm{m}$

Number of droplets converted from original drop = 64

Surface energies charge is

$\Delta(S E)=T \Delta A \ldots$ (i) $[\Delta A \rightarrow$ change in area

$ \begin{aligned} T & \rightarrow \text { surface tension } \\ S E & \rightarrow \text { surface energy }] \\ \Delta A & =\Delta_{\text {final }}-\Delta_{\text {initial }} \\ A_{\text {initial }} & =4 \pi r^2 \quad[\text { Area of sphere }] \\ A_{\text {inital }} & =4 \pi r_1^2 \times 64 \\ & {\left[r_1=\text { new radius of drops }\right] } \end{aligned} $

Volume will remains constant

$ \begin{aligned} V_{\text {initial }} & =V_{\text {final }} \\ \Rightarrow \quad \frac{4}{3} \pi r^3 & =\frac{4}{3} \pi r_1^3 \times 64 \\ \left(r_1=\frac{r}{4}\right) & \\ \Delta A_{\text {final }}= & 10^{-7} \times 64=64 \times 10^{-7} \mathrm{~m}^2 \end{aligned} $

$ \begin{aligned} \Delta A_{\text {initial }} & =4 \pi\left(4 r_1\right)^2=16 \times 4 \pi r_1^2 \\ & =16 \times 10^{-7} \mathrm{~m}^2 \\ \Delta A & =\left[\Delta A_{\text {final }}-\Delta A_{\text {initial }}\right] \\ \Delta(S E) & =0.07 \times\left[64 \times 10^{-7}-16 \times 10^{-7}\right] \\ \Delta(S E) & =336 \times 10^{-9} \mathrm{~J} \end{aligned} $

Given,

Total volume of vessel $=15 \mathrm{~L}$

5 L water in $t_1$ time.

Next 5 L water in $t_2$ time.

Last 5 L water in $t_3$ time.

Velocity of efflux is given by

$ v=\sqrt{2 g h} $

Let, For first 5 L initial height is $h_1$

For next 5 L height is $h_2$

For last 5 L height is $h_3$

$ \begin{array}{ll} \text { So, } & h_1>h_2>h_3 \\ \therefore & v_1>v_2>v_3 \end{array} $

As water flow out, the height of water level also decreases. This results in a decrease in water flow. i.e. rate of water level decrease with time.

Therefore $t_1

Given,

mass of cylinder $=m$

density of cylinder $=\rho$

vessel cross-section area $=A$

density of liquid $=\sigma$

$ \sigma>\rho $

The increase in pressure $\Delta p=\frac{\Delta F}{\text { Area }}$

$\Delta F=$ Buoyant force $=V g \sigma$

$ \begin{aligned} & \mathrm{V}=\text { Volume of cylinder }=\frac{M}{\rho} \\ & \Delta p=\frac{\frac{M}{\rho} \times \sigma \times g}{A} \Rightarrow \Delta p=\frac{M \sigma g}{\rho A} \end{aligned} $

Angle of contact is one of the characteristic of interaction of a liquid with a solid body.

So, it does not degreed on inclination or orientation of the road or capillary tube. So, angle of contact will be $120^{\circ} \mathrm{C}$

Let atmosphere pressure be

$ p_0=1.05 \times 10^5 \mathrm{~Pa} $

Pressure below 10 m in a well will be

$ \begin{aligned} p & =p_0+\rho g h \\ & =1.05 \times 10^5+1000 \times 10 \times 10 \\ & =1.99 \times 10^5 \\ & \approx 2 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \end{aligned} $

Given, radius $R=0.05 \mathrm{~m}$

Speed of rotation, $\omega=20 \mathrm{rad} \mathrm{s}^{-1}$

When the liquid is rotated it reaches a certain height $h$. At this point

$ \begin{aligned} h \rho g & =\frac{1}{2} \rho(r \omega)^2 \\ \Rightarrow \quad h & =\frac{\omega^2 r^2}{2 g}=\frac{(10)^2 \times(0.05)^2}{2 \times(10)} \\ & =\frac{100 \times 0.0025}{2 \times 10}=125 \times 10^{-4} \mathrm{~m} \end{aligned} $

Given height, $h=7 \mathrm{~cm}$,

$ =7 \times 10^{-2} \mathrm{~m} $

Surface tension of water $T$ or $S=0.07 \mathrm{Nm}^{-1}$

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

Surface tension of water membrane on the bottom larger of the water will be,

$ T=\frac{F}{2 \pi r}=\frac{\rho g h \times \pi r^2}{2 \pi r}=\frac{\rho g h r}{2} $

Where, $r$ is the radius of hole, $r=\frac{2 T}{\rho g h}$

$ \begin{aligned} & =\frac{0.07 \times 2}{0.07 \times 1000 \times 10} \\ & =2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm} \end{aligned} $

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.