Fluid Mechanics

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:

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,

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

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

$=2 \mathrm{~g}$

Using Pascal's law;

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

From Eq. (i) we have

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

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

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

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

Therefore, no option is correct.

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

Using the principle of hydraulic lifts, we have:

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

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

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

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

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

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

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

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

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

The given situation is shown below

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

By the principle of continuity,

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

According to Bernoulli's theorem,

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

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

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

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

As we know that,

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

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

Given,

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

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

Let specific gravity of glass $=\rho$

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

As we know that,

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

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

$\therefore$ Water does not wet glass.

As we know that,

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

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

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

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

$\therefore$ Option (c) is incorrect.

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

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

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

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

Let radius of capillary tube be $R$.

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

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

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

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

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

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

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

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

Substituting in Eq. (i), we get

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

If force is same.

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

If pressure is least, then velocity is highest.