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,

$ \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,

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.

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.

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} $

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} $