Mechanical Properties of Fluids — Competition Zone

A U-shaped tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is:

1. 650 kg m^-3
2. 425 kg m^-3
3. 800 kg m^-3
4. 928 kg m^-3.

Answer

928 kg m^-3

Reason — Given,

• Water rises by 65 mm from its original level
• The oil stands 10 mm above the water level on the other side
• Density of water, ρ_w = 1000 kg m^-3

Since the water in one limb rises by 65 mm, the water in the other limb falls by 65 mm. Hence, measured from the lowest common level of the two liquids,

$\text{height of the water column, h}_w = 65 + 65 = 130\ \text{mm}$

$\text{height of the oil column, h}_{oil} = 130 + 10 = 140\ \text{mm}$

The pressures at the common level in the two limbs must be equal,

$\text h_{oil} \times \rho_{oil} \times \text g = \text h_w \times \rho_w \times \text g$

Substituting the values,

$140 \times \rho_{oil} = 130 \times \rho_w$

$\rho_{oil} = \dfrac{13}{14} \times 1000 \\[1em] = 928\ \text{kg m}^{-3}$

The mass of a hydrogen molecule is 3.32 × 10^-27 kg. If 10²³ hydrogen molecules strike, per second, a fixed wall of area 2 cm² at an angle of 45° to the normal, and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly:

1. 2.35 × 10² N/m²
2. 4.70 × 10² N/m²
3. 2.35 × 10³ N/m²
4. 4.70 × 10³ N/m².

Answer

2.35 × 10³ N/m²

Reason — Given,

• Mass of a hydrogen molecule, m = 3.32 × 10^-27 kg
• Number of molecules striking per second, n = 10²³
• Area of the wall, A = 2 cm² = 2 × 10^-4 m²
• Speed of the molecules, v = 10³ m s^-1
• Angle with the normal, θ = 45°

Since the collision is elastic, the component of velocity parallel to the wall is unchanged, while the component normal to the wall (v cos 45°) is reversed. Hence the change in momentum of each molecule is

$\Delta \text p = 2\text mv\cos 45^\circ$

The force on the wall is the rate of change of momentum of all the n molecules striking per second,

$\text F = \dfrac{\text{dp}}{\text{dt}} = 2\text n\text mv\cos 45^\circ$

Therefore, the pressure on the wall is

$\text P = \dfrac{\text F}{\text A} = \dfrac{2\text n\text mv\cos 45^\circ}{\text A}$

Substituting the values,

$\text P = \dfrac{2 \times 10^{23} \times (3.32 \times 10^{-27}) \times 10^{3} \times \dfrac{1}{\sqrt 2}}{2 \times 10^{-4}} \\[1em] = \dfrac{0.4696}{2 \times 10^{-4}}$

$= 2.35 \times 10^{3}\ \text{N m}^{-2}$

A submarine experiences a pressure of 5.05 × 10⁶ Pa at a depth of d₁ in a sea. When it goes further to a depth of d₂, it experiences a pressure of 8.08 × 10⁶ Pa, then d₂ − d₁ is approximately (density of water = 10³ kg m^-3 and g = 10 m/s²) :

1. 500 m
2. 400 m
3. 600 m
4. 300 m.

Answer

300 m

Reason — Given,

• Pressure at depth d₁, p₁ = 5.05 × 10⁶ Pa
• Pressure at depth d₂, p₂ = 8.08 × 10⁶ Pa
• Density of water, ρ = 10³ kg m^-3
• g = 10 m s^-2

The pressure at a depth d below the free surface is

$\text p = \text p_0 + \text d\rho \text g$

where p₀ is the atmospheric pressure. Hence,

$\text p_1 = \text p_0 + \text d_1\rho \text g \quad \text{and} \quad \text p_2 = \text p_0 + \text d_2\rho \text g$

Subtracting,

$\text p_2 - \text p_1 = (\text d_2 - \text d_1)\rho \text g$

$\text d_2 - \text d_1 = \dfrac{\text p_2 - \text p_1}{\rho \text g}$

Substituting the values,

$\text d_2 - \text d_1 = \dfrac{(8.08 \times 10^{6}) - (5.05 \times 10^{6})}{10^{3} \times 10} = \dfrac{3.03 \times 10^{6}}{10^{4}} \\[1em] = 303\ \text m \simeq 300\ \text m$

A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:

1. 0.6
2. 0.8
3. 0.7
4. 0.5.

Answer

0.6

Reason — Given,

• In the first case, the volume submerged in water is $\dfrac{4}{5}$V
• In the second case, half the volume is in water and half in oil

First condition : By the law of floatation, the weight of the block equals the weight of the water displaced,

$\text V\rho_b\text g = \dfrac{4}{5}\text V\rho_w\text g \quad \Rightarrow \quad \dfrac{\rho_b}{\rho_w} = \dfrac{4}{5} \qquad \ldots(\text i)$

Second condition : The weight of the block is now balanced by the weight of the oil displaced together with the weight of the water displaced,

$\text V\rho_b\text g = \dfrac{\text V}{2}\rho_0\text g + \dfrac{\text V}{2}\rho_w\text g$

$\rho_b = \dfrac{\rho_0}{2} + \dfrac{\rho_w}{2} \quad \Rightarrow \quad 2\rho_b = \rho_0 + \rho_w$

Dividing throughout by ρ_w,

$\dfrac{2\rho_b}{\rho_w} = \dfrac{\rho_0}{\rho_w} + 1$

Substituting from equation (i),

$2 \times \dfrac{4}{5} = \dfrac{\rho_0}{\rho_w} + 1 \quad \Rightarrow \quad \dfrac{8}{5} = \dfrac{\rho_0}{\rho_w} + 1$

$\dfrac{\rho_0}{\rho_w} = \dfrac{8}{5} - 1 = \dfrac{3}{5} = 0.6$

The area of cross section of the rope used to lift a load by a crane is 2.5 × 10^-4 m². The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons. The required area of cross section of the rope should be : (Take g = 10 ms^-2)

1. 6.25 × 10^-4 m²
2. 10 × 10^-4 m²
3. 1 × 10^-4 m²
4. 1.67 × 10^-4 m².

Answer

6.25 × 10^-4 m²

Reason — Given,

• Initial area of cross-section, A₁ = 2.5 × 10^-4 m²
• Initial lifting capacity, W₁ = 10 metric tons
• Final lifting capacity, W₂ = 25 metric tons

The rope can bear only a fixed maximum stress, that is, a fixed force per unit area. Hence the load which can be lifted is directly proportional to the area of cross-section of the rope,

$\dfrac{\text W_1}{\text A_1} = \dfrac{\text W_2}{\text A_2}$

$\text A_2 = \dfrac{\text W_2}{\text W_1} \times \text A_1$

Substituting the values,

$\text A_2 = \dfrac{25}{10} \times (2.5 \times 10^{-4}) \\[1em] = 6.25 \times 10^{-4}\ \text m^2$

A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to:

1. r³
2. r⁴
3. r⁵
4. r².

Answer

r⁵

Reason — The rate of production of heat is equal to the power lost against the viscous force,

$\dfrac{\text{dQ}}{\text{dt}} = \text F_v \times v$

By Stokes' law, F_v = 6πηrv, so

$\dfrac{\text{dQ}}{\text{dt}} = (6\pi \eta \text{rv}) \times v = 6\pi \eta \text rv^2$

At the terminal velocity,

$v = \dfrac{2}{9}\dfrac{\text r^2(\rho - \sigma)\text g}{\eta} \quad \Rightarrow \quad v \propto \text r^2$

Therefore,

$\dfrac{\text{dQ}}{\text{dt}} \propto \text rv^2 \propto \text r \times (\text r^2)^2 = \text r^5$

A solid sphere of radius R acquires a terminal velocity v₁ when falling (due to gravity) through a viscous fluid having co-efficient of viscosity η. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity v₂, when falling through the same fluid, the ratio (v₁/v₂) equals:

1. 9

2. $\dfrac{1}{27}$

3. $\dfrac{1}{9}$

Answer

9

Reason — Let r be the radius of each of the 27 small spheres. Since the total volume remains unchanged,

$\dfrac{4}{3}\pi \text R^3 = 27 \times \dfrac{4}{3}\pi \text r^3$

$\text R^3 = 27\text r^3 \quad \Rightarrow \quad \text R = 3\text r$

By Stokes' law, the terminal velocity is

$v_T = \dfrac{2}{9}\dfrac{\text r^2(\rho - \sigma)\text g}{\eta}$

Since the material of the sphere and the fluid remain the same, all the other parameters are constant and

$v_T \propto (\text{radius})^2$

Therefore,

$\dfrac{v_1}{v_2} = \dfrac{\text R^2}{\text r^2} = \dfrac{(3\text r)^2}{\text r^2} = 9$

Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of the tap is 10^-4 m². Assume that the pressure is constant throughout the stream of water and that the flow is stream-lined. The cross-sectional area of the stream, 0.15 m below the tap would be : [Take g = 10 m/s²]

1. 2 × 10^-5 m²
2. 1 × 10^-5 m²
3. 5 × 10^-4 m²
4. 5 × 10^-5 m².

Answer

5 × 10^-5 m²

Reason — Given,

• Initial speed of water, v₁ = 1.0 m s^-1
• Area of cross-section of the tap, A₁ = 10^-4 m²
• Vertical distance fallen, h = 0.15 m
• g = 10 m s^-2

Since the pressure is constant throughout the stream, the water falls freely under gravity. Using the equation of motion v² = u² + 2gh, the speed at a depth of 0.15 m is

$v_2^2 = v_1^2 + 2\text{gh} = (1)^2 + 2 \times 10 \times 0.15 \\[1em] = 1 + 3 = 4$

$v_2 = 2\ \text{m s}^{-1}$

By the equation of continuity, at the two points P and Q,

$\text A_1v_1 = \text A_2v_2 \quad \Rightarrow \quad \text A_2 = \dfrac{\text A_1v_1}{v_2}$

Substituting the values,

$\text A_2 = \dfrac{10^{-4} \times 1}{2} \\[1em] = 5 \times 10^{-5}\ \text m^2$

A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

1. $\rho v^2$

2. $\dfrac{1}{2}\rho v^2$

3. $\dfrac{1}{4}\rho v^2$

4. $\dfrac{3}{4}\rho v^2$.

Answer

$\dfrac{3}{4}\rho v^2$

Reason — The mass of liquid flowing per unit time through a pipe of area of cross-section A is

$\dfrac{\text{dm}}{\text{dt}} = \text Av\rho$

For the 25% which loses all its momentum : the change of velocity is v, so the rate of change of momentum is

$\dfrac{\text{dp}_1}{\text{dt}} = \dfrac{1}{4}\left(\dfrac{\text{dm}}{\text{dt}}\right)v = \dfrac{1}{4}\text Av^2\rho$

For the 25% which comes back with the same speed : the net change of velocity is 2v, so

$\dfrac{\text{dp}_2}{\text{dt}} = \dfrac{1}{4}\left(\dfrac{\text{dm}}{\text{dt}}\right) \times 2v = \dfrac{1}{2}\text Av^2\rho$

The 50% which passes through unaffected contributes nothing. Hence the net force on the mesh is the sum of the two rates of change of momentum, and the resultant pressure is

$\text P = \dfrac{\dfrac{\text{dp}_1}{\text{dt}} + \dfrac{\text{dp}_2}{\text{dt}}}{\text A} = \dfrac{\dfrac{1}{4}\text A v^2\rho + \dfrac{1}{2}\text A v^2\rho}{\text A}$

$\text P = \dfrac{3}{4}\rho v^2$

Water flows into a large tank with flat bottom at the rate of 10^-4 m³/s. Water is also leaking out of a hole of area 1 cm² at its bottom. If the height of the water in the tank remains steady, then this height is:

1. 4 cm
2. 2.9 cm
3. 5.1 cm
4. 1.7 cm.

Answer

5.1 cm

Reason — Given,

• Rate of inflow of water = 10^-4 m³ s^-1
• Area of the hole, A = 1 cm² = 10^-4 m²
• g = 9.8 m s^-2

Since the water level in the tank remains steady, the rate of inflow of water must be equal to the rate of outflow of water,

$\text{rate of inflow} = \text{area of the hole} \times \text{velocity of efflux}$

By Torricelli's theorem, the velocity of efflux is $\sqrt{2\text{gh}}$, where h is the height of water above the orifice. Therefore,

$10^{-4} = 10^{-4} \times \sqrt{2\text{gh}}$

$\sqrt{2\text{gh}} = 1 \quad \Rightarrow \quad 2\text{gh} = 1$

$\text h = \dfrac{1}{2\text g} = \dfrac{1}{2 \times 9.8} = \dfrac{100}{19.6}\ \text{cm} \\[1em] = 5.1\ \text{cm}$

Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm then the Reynold's number for the flow is of the order of : (Density of water = 1000 kg/m³, co-efficient of viscosity of water = 1 Pa-s)

1. 10³
2. 10⁴
3. 10²
4. 10⁶.

Answer

10⁴

Reason — Given,

• Volume flow rate, $\dfrac{\text V}{\text t} = 100$ litres/min $= \dfrac{100 \times 10^{-3}}{60}$ m³ s^-1
• Radius of the pipe, r = 5 cm = 5 × 10^-2 m
• Density of water, ρ = 1000 kg m^-3
• Coefficient of viscosity of water, η = 1 × 10^-3 Pa-s

Reynold's number is given by

$\text R_e = \dfrac{\rho v\text D}{\eta}$

The velocity of flow is the volume flow rate divided by the area of cross-section,

$v = \dfrac{\text V/\text t}{\text A} = \dfrac{\text V}{\pi \text r^2\text t}$

Substituting D = 2r,

$\text R_e = \dfrac{\rho (2\text r)}{\eta} \times \dfrac{\text V}{\pi \text r^2\text t} = \dfrac{2\rho \text V}{\pi \eta \text{rt}}$

Substituting the values,

$\text R_e = \dfrac{2 \times 1000 \times \dfrac{100 \times 10^{-3}}{60}}{3.14 \times (1 \times 10^{-3}) \times (5 \times 10^{-2})} \\[1em] = \dfrac{3.333}{1.57 \times 10^{-4}}$

$= 2.12 \times 10^{4}$

Hence, the Reynold's number is of the order of 10⁴.

Note: The value of the coefficient of viscosity of water printed in the question, 1 Pa-s, is not the correct value for water. The textbook's own solution uses η = 1 × 10^-3 Pa-s, which is the standard value for water, and this has been used here.

A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:

1. 8.9 × 10^-6 m³/s
2. 2.23 × 10^-6 m³/s
3. 6.4 × 10^-6 m³/s
4. 12.6 × 10^-6 m³/s.

Answer

12.6 × 10^-6 m³/s

Reason — Given,

• Area of the hole, A = 2 mm² = 2 × 10^-6 m²
• Height of the tank, h = 2 m
• g = 10 m s^-2

Since the hole is near the bottom of a fully filled tank, the height of water above it is 2 m. By Torricelli's theorem, the velocity of efflux is

$v = \sqrt{2\text{gh}} = \sqrt{2 \times 10 \times 2} = \sqrt{40}\ \text{m s}^{-1}$

The rate of flow of water is

$\text Q = \text A \times v$

Substituting the values,

$\text Q = (2 \times 10^{-6}) \times \sqrt{40} \\[1em] = (2 \times 10^{-6}) \times 6.32$

$= 12.6 \times 10^{-6}\ \text m^3\text s^{-1}$

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is:

1. C
2. D
3. A
4. B.

Answer

B

Reason — When the ball is dropped in a viscous liquid, its speed is initially small, so the viscous force on it is small and the ball falls with a nearly uniform acceleration. Hence the speed increases in the beginning.

As time passes, the viscous drag acting on the ball increases with its increasing speed. The effective driving force is thereby reduced, so the rate of increase of speed becomes smaller and smaller.

Finally a stage comes when the upward viscous force becomes equal to the effective downward force. The net force then becomes zero and the ball moves with a constant speed, called the terminal velocity.

Hence, the required graph is the one in which the speed first increases and then becomes constant, which is the curve B.

The ratio of surface tensions of mercury and water is given as 7.5 while the ratio of their densities is 13.6. Their contact angles with glass are close to 135° and 0° respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r₁, while water rises by the same amount h in a capillary tube of radius r₂. The ratio (r₁/r₂) is then close to:

1. $\dfrac{3}{5}$

2. $\dfrac{2}{3}$

3. $\dfrac{2}{5}$

4. $\dfrac{4}{5}$.

Answer

$\dfrac{2}{5}$

Reason — Given,

• $\dfrac{\text T_{Hg}}{\text T_w} = 7.5$, $\dfrac{\rho_{Hg}}{\rho_w} = 13.6$
• θ_Hg = 135°, θ_w = 0°, and h_Hg = h_w = h in magnitude

By the ascent formula, the height of the liquid column in a capillary tube is

$\text h = \dfrac{2\text T\cos \theta}{\text r\rho \text g}$

Since the magnitudes of the two heights are the same,

$\dfrac{2\text T_{Hg}\cos \theta_{Hg}}{\rho_{Hg}\text g\text r_1} = \dfrac{2\text T_w\cos \theta_w}{\rho_w\text g\text r_2}$

Rearranging,

$\dfrac{\text r_1}{\text r_2} = \dfrac{\text T_{Hg}}{\text T_w} \times \dfrac{\cos \theta_{Hg}}{\cos \theta_w} \times \dfrac{\rho_w}{\rho_{Hg}}$

Substituting the values, with cos 135° = $-\dfrac{1}{\sqrt 2}$ and cos 0° = 1,

$\dfrac{\text r_1}{\text r_2} = 7.5 \times \left(-\dfrac{1}{\sqrt 2}\right) \times \dfrac{1}{13.6} \\[1em] = -0.39 \simeq -\dfrac{2}{5}$

The negative sign only indicates that the mercury gets depressed while water rises. Taking the magnitude,

$\dfrac{\text r_1}{\text r_2} = \dfrac{2}{5}$

If m is the mass of water that rises in a capillary tube of radius r, then mass of water which will rise in a capillary tube of radius 2r is:

1. 2m

2. 4m

3. $\dfrac{m}{2}$

4. m.

Answer

2m

Reason — By the ascent formula, the height of the liquid column in a capillary tube is

$\text h = \dfrac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text h \propto \dfrac{1}{\text r}$

So, when the radius is doubled, the height of rise becomes half, that is, $\text h' = \dfrac{\text h}{2}$.

The mass of the liquid inside the capillary is

$\text m = \rho \times \text{volume} = \rho \pi \text r^2\text h$

Therefore, for the tube of radius 2r,

$\dfrac{\text m'}{\text m} = \dfrac{\rho \pi (2\text r)^2\text h'}{\rho \pi \text r^2\text h} = \dfrac{(2\text r)^2\left(\dfrac{\text h}{2}\right)}{\text r^2\text h}$

$= \dfrac{4\text r^2}{2\text r^2} = 2 \quad \Rightarrow \quad \text m' = 2\text m$

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10^-2 N/m. The pressure inside the bubble equals at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s², density of water = 10³ kg/m³, the value of Z₀ is:

1. 10 cm
2. 1 cm
3. 0.5 cm
4. 100 cm.

Answer

1 cm

Reason — Given,

• Radius of the soap bubble, R = 1 mm = 1 × 10^-3 m
• Surface tension of the detergent solution, T = 2.5 × 10^-2 N m^-1
• Density of water, ρ = 10³ kg m^-3
• g = 10 m s^-2

A soap bubble has two free surfaces, so the pressure inside it is

$\text P_{in} = \text P_0 + \dfrac{4\text T}{\text R}$

The pressure at a depth Z₀ below the free surface of water is

$\text P = \text P_0 + \rho \text{gZ}_0$

Since the two pressures are equal,

$\text P_0 + \dfrac{4\text T}{\text R} = \text P_0 + \rho \text{gZ}_0$

$\text Z_0 = \dfrac{4\text T}{\text R\rho \text g}$

Substituting the values,

$\text Z_0 = \dfrac{4 \times (2.5 \times 10^{-2})}{(1 \times 10^{-3}) \times 10^{3} \times 10} = \dfrac{0.1}{10} \\[1em] = 0.01\ \text m = 1\ \text{cm}$

A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be : (Given π = 3.14)

1. 8.5 × 10^-4 J
2. 8.2 × 10^-4 J
3. 7.5 × 10^-4 J
4. 5.3 × 10^-4 J.

Answer

7.5 × 10^-4 J

Reason — Given,

• Radius of the drop, R = 1 cm = 1 × 10^-2 m
• Number of droplets, n = 729
• Surface tension of water, T = 75 dyne/cm = 0.075 N m^-1

Since the total volume remains unchanged,

$729 \times \dfrac{4}{3}\pi \text r^3 = \dfrac{4}{3}\pi \text R^3$

$\text R^3 = 729\text r^3 \quad \Rightarrow \quad \text R = 9\text r \qquad \ldots(\text i)$

The gain in surface energy is the surface tension multiplied by the increase in the total surface area,

$\Delta \text U = \text T \times \Delta \text A = \text T[729 \times 4\pi \text r^2 - 4\pi \text R^2]$

Substituting $\text r = \dfrac{\text R}{9}$ from equation (i),

$\Delta \text U = 4\pi \text T\left[729 \times \dfrac{\text R^2}{81} - \text R^2\right] = 4\pi \text T[9\text R^2 - \text R^2]$

$= 4\pi \text T \times 8\text R^2$

Substituting the values,

$\Delta \text U = 4 \times 3.14 \times 0.075 \times 8 \times (1 \times 10^{-2})^2 \\[1em] = 0.942 \times 8 \times 10^{-4}$

$= 7.5 \times 10^{-4}\ \text J$

If a soap bubble expands, the pressure inside the bubble:

1. remains the same
2. is equal to the atmospheric pressure
3. decreases
4. increases.

Answer

decreases

Reason — For a soap bubble, the excess pressure inside it is

$\text P_{in} - \text P_{out} = \dfrac{4\text T}{\text R}$

When the bubble expands, its radius R increases, so the quantity $\dfrac{4\text T}{\text R}$ decreases.

Since the pressure outside the bubble is the atmospheric pressure and remains constant, the pressure inside the bubble must also decrease.

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If the surface tension of water is 0.07 Nm^-1, then the excess force required to take it away from the surface is:

1. 19.8 N
2. 198 N
3. 19.8 mN
4. 99 N.

Answer

19.8 mN

Reason — Given,

• Radius of the disc, R = 4.5 cm = 4.5 × 10^-2 m
• Surface tension of water, T = 0.07 N m^-1

The surface tension acts along the circumference of the circular disc which is in contact with the water surface. Hence the length of the line of contact is

$\text l = 2\pi \text R$

The excess force required to take the disc away from the surface is

$\text F = \text T \times 2\pi \text R$

Substituting the values, with $\pi = \dfrac{22}{7}$,

$\text F = 0.07 \times 2 \times \dfrac{22}{7} \times (4.5 \times 10^{-2}) \\[1em] = 0.07 \times 0.2829$

$= 1.98 \times 10^{-2}\ \text N = 19.8\ \text{mN}$

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the Z-direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ₀ (θ₀ << 1) with the X-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is:

$\theta(x) = \sin \theta(x) = \tan \theta(x) = \dfrac{dy}{dx}$

[take θ(x) = sin θ(x) = tan θ(x), g is the acceleration due to gravity]

1. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{\text S}x$

2. $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{\text S}y$

3. $\dfrac{d^2y}{dx^2} = \sqrt{\dfrac{\rho g}{\text S}}$

4. $\dfrac{dy}{dx} = \sqrt{\dfrac{\rho g}{\text S}}x$

Answer

$\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{\text S}y$

Reason — Consider a small element of the liquid surface of width dx, taking unit length d in the Z-direction.

The forces of surface tension acting on the two edges of this element make angles θ and (θ + dθ) with the horizontal. The net upward force due to surface tension is

$\text F_y = [\text S\sin(\theta + \text d\theta) - \text S\sin \theta]\text d$

This net upward force supports the weight of the liquid element, whose height is y and width dx, so

$[\text S\sin(\theta + \text d\theta) - \text S\sin \theta]\text d = \text{mg} = \rho (\text y\text{dx}\text d)\text g$

Since the angle is small, sin θ ≈ θ, and therefore

$\text S\text d\theta = \rho \text{gy}\text{dx} \quad \Rightarrow \quad \dfrac{\text d\theta}{\text{dx}} = \dfrac{\rho \text{gy}}{\text S} \qquad \ldots(\text i)$

Also, taking $\tan \theta = \dfrac{\text{dy}}{\text{dx}}$ and differentiating with respect to x,

$\sec^2\theta\dfrac{\text d\theta}{\text{dx}} = \dfrac{\text d^2\text y}{\text{dx}^2} \qquad \ldots(\text{ii})$

Since θ is small, cos θ ≈ 1 and hence sec²θ ≈ 1. Substituting the value of $\dfrac{\text d\theta}{\text{dx}}$ from equation (i) into equation (ii),

$\dfrac{\text d^2\text y}{\text{dx}^2} = \dfrac{\rho \text g}{\text S}\text y$

Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u₀. Which of the following statements is (are) true?

1. The resistive force of liquid on the plate is inversely proportional to h
2. The resistive force of liquid on the plate is independent of the area of the plate
3. The tangential (shear) stress on the floor of the tank increases with u₀
4. The tangential (shear) stress on the plate varies linearly with the viscosity η of the liquid.

Answer

1. The resistive force of liquid on the plate is inversely proportional to h

2. The tangential (shear) stress on the floor of the tank increases with u₀

3. The tangential (shear) stress on the plate varies linearly with the viscosity η of the liquid.

Reason — Given,

• Height of the liquid in the tank = h
• Velocity with which the plate is pulled = u₀
• Coefficient of viscosity of the liquid = η

By Newton's law of viscosity, the viscous force on the plate of area A is

$\text F_v = \eta \text A\dfrac{\text{dv}}{\text{dz}}$

Since the height h of the liquid in the tank is very small, the velocity gradient may be taken as uniform,

$\dfrac{\text{dv}}{\text{dz}} = \dfrac{\Delta v}{\Delta \text z} = \dfrac{\text u_0}{\text h}$

Therefore,

$\text F_v = \eta \text A\left(\dfrac{\text u_0}{\text h}\right)$

Option 1 : From the above relation, $\text F_v \propto \dfrac{1}{\text h}$. Hence option 1 is correct.

Option 2 : The relation also shows that F_v ∝ A, so the resistive force is not independent of the area of the plate. Hence option 2 is incorrect.

Option 3 : The tangential (shear) stress is

$\dfrac{\text F_v}{\text A} = \dfrac{\eta \text u_0}{\text h}$

so the shear stress is directly proportional to u₀ and increases with it. Hence option 3 is correct.

Option 4 : The same relation shows that $\dfrac{\text F_v}{\text A} \propto \eta$, that is, the shear stress varies linearly with the viscosity of the liquid. Hence option 4 is correct.

An ideal gas of density ρ = 0.2 kg m^-3 enters a chimney of height h at the rate of α = 0.8 kg s^-1 from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is A₁ = 0.1 m² and the upper end is A₂ = 0.4 m². The pressure and the temperature of the gas at the lower end are 600 Pa and 300 K, respectively, while its temperature at the upper end is 150 K. The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take g = 10 ms^-2 and the ratio of specific heats of the gas γ = 2. Ignore atmospheric pressure.

Which of the following statement(s) is(are) correct?

1. The pressure of the gas at the upper end of the chimney is 300 Pa.
2. The velocity of the gas at the lower end of the chimney is 40 ms^-1 and at the upper end is 20 ms^-1.
3. The height of the chimney is 590 m.
4. The density of the gas at the upper end is 0.05 kg m^-3.

Answer

2. The velocity of the gas at the lower end of the chimney is 40 ms^-1 and at the upper end is 20 ms^-1.

3. The height of the chimney is 590 m.

Reason — Given,

• Density of the gas at the lower end, ρ₁ = 0.2 kg m^-3
• Rate of mass flow, α = 0.8 kg s^-1
• A₁ = 0.1 m², A₂ = 0.4 m²
• P₁ = 600 Pa, T₁ = 300 K, T₂ = 150 K
• γ = 2, g = 10 m s^-2

Density at the upper end : For an adiabatic process, $\dfrac{\rho_1}{\rho_2} = \left(\dfrac{\text T_1}{\text T_2}\right)^{\frac{1}{\gamma - 1}}$, so

$\rho_2 = \rho_1\left(\dfrac{\text T_2}{\text T_1}\right)^{\frac{1}{\gamma - 1}} = 0.2 \times \left(\dfrac{150}{300}\right)^{1} \\[1em] = 0.1\ \text{kg m}^{-3}$

Since this is 0.1 kg m^-3 and not 0.05 kg m^-3, option 4 is incorrect.

Pressure at the upper end : For an adiabatic process,

$\dfrac{\text P_2}{\text P_1} = \left(\dfrac{\rho_2}{\rho_1}\right)^{\gamma} \quad \Rightarrow \quad \text P_2 = 600\left(\dfrac{1}{2}\right)^{2}$

$\text P_2 = 150\ \text{Pa}$

Since this is 150 Pa and not 300 Pa, option 1 is incorrect.

Velocities : The rate of mass flow is α = ρ A v, so the volume flow rate at the lower end is

$\dfrac{\text V_1}{\text t} = \dfrac{\alpha}{\rho_1} = \dfrac{0.8}{0.2} = 4\ \text m^3\text s^{-1}$

$v_1 = \dfrac{\text V_1/\text t}{\text A_1} = \dfrac{4}{0.1} = 40\ \text{m s}^{-1}$

At the upper end,

$v_2 = \dfrac{\alpha}{\rho_2\text A_2} = \dfrac{0.8}{0.1 \times 0.4} = 20\ \text{m s}^{-1}$

Hence option 2 is correct.

Height of the chimney : Applying Bernoulli's theorem between the two ends,

$\dfrac{1}{2}\rho_1v_1^2 + \text P_1 = \dfrac{1}{2}\rho_2v_2^2 + \text P_2 + \rho_2\text{gh}$

Substituting the values,

$\dfrac{1}{2}(0.2)(1600) + 600 = \dfrac{1}{2}(0.1)(400) + 150 + (0.1)(10)\text h \\[1em] 160 + 600 = 20 + 150 + \text h$

$\text h = 760 - 170 = 590\ \text m$

Hence option 3 is correct.

A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is T. The angle of contact between water and the wall of the capillary tube is θ. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?

1. For a given material of the capillary tube, h decreases with increase in r
2. For a given material of the capillary tube, h is independent of T
3. If this experiment is performed in a lift going up with a constant acceleration, then h decreases
4. h is proportional to contact angle θ.

Answer

1. For a given material of the capillary tube, h decreases with increase in r

2. If this experiment is performed in a lift going up with a constant acceleration, then h decreases

Reason — By the ascent formula, the height to which water rises in a capillary tube is

$\text h = \dfrac{2\text T\cos \theta}{\text r\rho \text g}$

Option 1 : From the above relation, $\text h \propto \dfrac{1}{\text r}$, so for a given material of the tube (θ fixed), h decreases as r increases. Hence option 1 is correct.

Option 2 : The same relation gives h ∝ T, so h is not independent of the surface tension. Hence option 2 is incorrect.

Option 3 : When the lift moves up with a constant acceleration a, the effective acceleration due to gravity becomes (g + a). Hence,

$\text h = \dfrac{2\text T\cos \theta}{\rho (\text g + \text a)\text r}$

Since the denominator increases, h decreases. Hence option 3 is correct.

Option 4 : The relation shows that h ∝ cos θ, and not h ∝ θ. Hence option 4 is incorrect.

A cylindrical capillary tube of 0.2 mm radius is made by joining two capillaries T₁ and T₂ of different materials having water contact angles of 0° and 60°, respectively. The capillary tube is dipped vertically in water in two different configurations, case I and II as shown in figure. Which of the following option(s) is(are) correct? [Surface tension of water = 0.075 N/m, density of water = 1000 kg/m³, take g = 10 m/s²]

1. For case I, if the joint is kept at 8 cm above the water surface, the height of water column in the tube will be 7.5 cm. (Neglect the weight of the water in the meniscus).
2. For case I, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be more than 8.75 cm. (Neglect the weight of the water in the meniscus).
3. The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus, will be different for both cases.
4. For case II, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be 3.75 cm. (Neglect the weight of the water in the meniscus).

Answer

1. For case I, if the joint is kept at 8 cm above the water surface, the height of water column in the tube will be 7.5 cm. (Neglect the weight of the water in the meniscus).

2. The correction in the height of water column raised in the tube, due to weight of water contained in the meniscus, will be different for both cases.

3. For case II, if the capillary joint is 5 cm above the water surface, the height of water column raised in the tube will be 3.75 cm. (Neglect the weight of the water in the meniscus).

Reason — Given,

• Radius of the capillary tube, r = 0.2 mm = 0.2 × 10^-3 m
• Contact angle for T₁, θ₁ = 0°, and for T₂, θ₂ = 60°
• Surface tension of water, T = 0.075 N m^-1
• Density of water, ρ = 1000 kg m^-3, g = 10 m s^-2

If only a single material were used for the whole tube, the height of rise would be

For θ₁ = 0° :

$\text h_1 = \dfrac{2\text T\cos \theta_1}{\text r\rho \text g} = \dfrac{2 \times 0.075 \times \cos 0^\circ}{(0.2 \times 10^{-3}) \times 1000 \times 10} \\[1em] = \dfrac{0.15}{2} = 0.075\ \text m = 7.5\ \text{cm}$

For θ₂ = 60° :

$\text h_2 = \dfrac{2\text T\cos \theta_2}{\text r\rho \text g} = \dfrac{2 \times 0.075 \times 0.5}{2} \\[1em] = 0.0375\ \text m = 3.75\ \text{cm}$

Option 1 : In case I the lower capillary is T₂ and the joint is at 8 cm, which is above 7.5 cm. Hence the water rises only to 7.5 cm, which is within the lower tube. Option 1 is correct.

Option 2 : If the joint is at 5 cm, the water rises up to the joint and then enters the upper tube, where the meniscus changes its radius of curvature. Since the tube above cannot support a column higher than the value fixed by its own contact angle, the height cannot be more than 8.75 cm. Hence option 2 is incorrect.

Option 3 : The weight of water contained in the meniscus depends upon the shape of the meniscus, which is decided by the angle of contact. Since the angles of contact are different in the two cases, the correction will also be different. Option 3 is correct.

Option 4 : In case II the lower capillary is T₁ and the joint is at 5 cm. The water would rise to 7.5 cm in T₁ alone, so it reaches the joint and enters T₂, which can support a column of only 3.75 cm. Option 4 is correct.

A bubble has surface tension S. The ideal gas inside the bubble has ratio of specific heats $\gamma = \dfrac{5}{3}$. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is P_a1, the radius of the bubble is found to be r₁ and the temperature of the enclosed gas is T₁. When the atmospheric pressure is P_a2, the radius of the bubble and the temperature of the enclosed gas are r₂ and T₂, respectively. Which of the following statement(s) is(are) correct?

1. If the surface of the bubble is a perfect heat insulator, then

$\left(\dfrac{r_1}{r_2}\right)^5 = \dfrac{\text P_{a2} + \dfrac{2\text S}{r_2}}{\text P_{a1} + \dfrac{2\text S}{r_1}}$

2. If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.

3. If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then

$\left(\dfrac{r_1}{r_2}\right)^3 = \dfrac{\text P_{a2} + \dfrac{4\text S}{r_2}}{\text P_{a1} + \dfrac{4\text S}{r_1}}$

4. If the surface of the bubble is a perfect heat insulator, then

$\left(\dfrac{\text T_2}{\text T_1}\right)^{\frac{5}{2}} = \dfrac{\text P_{a2} + \dfrac{4\text S}{r_2}}{\text P_{a1} + \dfrac{4\text S}{r_1}}$

Answer

3. If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then

$\left(\dfrac{r_1}{r_2}\right)^3 = \dfrac{\text P_{a2} + \dfrac{4\text S}{r_2}}{\text P_{a1} + \dfrac{4\text S}{r_1}}$

4. If the surface of the bubble is a perfect heat insulator, then

$\left(\dfrac{\text T_2}{\text T_1}\right)^{\frac{5}{2}} = \dfrac{\text P_{a2} + \dfrac{4\text S}{r_2}}{\text P_{a1} + \dfrac{4\text S}{r_1}}$

Reason — Since it is a soap bubble, it has two free surfaces, so the pressure of the gas inside the bubble is

$\text P_1 = \text P_{a1} + \dfrac{4\text S}{\text r_1} \quad \text{and} \quad \text P_2 = \text P_{a2} + \dfrac{4\text S}{\text r_2}$

Options 1 and 4 (perfect heat insulator) : The process is adiabatic, so $\text P_1\text V_1^{\gamma} = \text P_2\text V_2^{\gamma}$. Since $\text V = \dfrac{4}{3}\pi \text r^3$,

$\left(\text P_{a1} + \dfrac{4\text S}{\text r_1}\right)\left(\dfrac{4}{3}\pi \text r_1^3\right)^{\frac{5}{3}} = \left(\text P_{a2} + \dfrac{4\text S}{\text r_2}\right)\left(\dfrac{4}{3}\pi \text r_2^3\right)^{\frac{5}{3}}$

$\left(\dfrac{\text r_1}{\text r_2}\right)^{5} = \dfrac{\text P_{a2} + \dfrac{4\text S}{\text r_2}}{\text P_{a1} + \dfrac{4\text S}{\text r_1}}$

Option 1 has 2S/r instead of 4S/r in the excess-pressure terms, so option 1 is incorrect.

Further, for an adiabatic process $\text T_1\text V_1^{\gamma - 1} = \text T_2\text V_2^{\gamma - 1}$, which gives

$\dfrac{\text T_2}{\text T_1} = \left(\dfrac{\text V_1}{\text V_2}\right)^{\gamma - 1} = \left(\dfrac{\text r_1}{\text r_2}\right)^{3\left(\frac{2}{3}\right)} = \left(\dfrac{\text r_1}{\text r_2}\right)^{2}$

Therefore,

$\left(\dfrac{\text T_2}{\text T_1}\right)^{\frac{5}{2}} = \left(\dfrac{\text r_1}{\text r_2}\right)^{5} = \dfrac{\text P_{a2} + \dfrac{4\text S}{\text r_2}}{\text P_{a1} + \dfrac{4\text S}{\text r_1}}$

Hence option 4 is correct.

Option 2 : If the surface of the bubble is a perfect heat insulator, the gas still does work on the surroundings as the bubble expands or contracts. Hence the total internal energy including the surface energy does not remain unchanged. Option 2 is incorrect.

Option 3 (perfect heat conductor) : Here the temperature remains constant, so the process is isothermal and Boyle's law applies, P₁V₁ = P₂V₂,

$\left(\text P_{a1} + \dfrac{4\text S}{\text r_1}\right)\left(\dfrac{4}{3}\pi \text r_1^3\right) = \left(\text P_{a2} + \dfrac{4\text S}{\text r_2}\right)\left(\dfrac{4}{3}\pi \text r_2^3\right)$

$\left(\dfrac{\text r_1}{\text r_2}\right)^{3} = \dfrac{\text P_{a2} + \dfrac{4\text S}{\text r_2}}{\text P_{a1} + \dfrac{4\text S}{\text r_1}}$

Hence option 3 is correct.

A table tennis ball has radius $\left(\dfrac{3}{2}\right) \times 10^{-2}$ m and mass $\left(\dfrac{22}{7}\right) \times 10^{-3}$ kg. It is slowly pushed down into a swimming pool to a depth of d = 0.7 m below the water surface and then released from rest. It emerges from the water surface at speed v, without getting wet and rises up to a height H. Which of the following option(s) is(are) correct? [Given: $\pi = \dfrac{22}{7}$, g = 10 ms^-2, density of water = 1 × 10³ kg m^-3, viscosity of water = 1 × 10^-3 Pa-s.]

1. The work done in pushing the ball to the depth d is 0.077 J.

2. If we neglect the viscous force in water, then the speed v = 7 m/s.

3. If we neglect the viscous force in water, then the height H = 1.4 m.

4. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $\dfrac{500}{9}$.

Answer

1. The work done in pushing the ball to the depth d is 0.077 J.

2. If we neglect the viscous force in water, then the speed v = 7 m/s.

Reason — Given,

• Radius of the ball, r = $\dfrac{3}{2} \times 10^{-2}$ m
• Mass of the ball, m = $\dfrac{22}{7} \times 10^{-3}$ kg
• Depth, d = 0.7 m; g = 10 m s^-2; ρ_w = 1 × 10³ kg m^-3

The volume of the ball is

$\text V = \dfrac{4}{3}\pi \text r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times \left(\dfrac{3}{2} \times 10^{-2}\right)^3$

Option 1 : The ball is pushed down slowly, so its kinetic energy remains zero and the total work done on it is zero,

$\text W_g + \text W_B + \text W_{ext} = 0$

Hence the external work done is the difference between the work done against the buoyant force and the work done by gravity,

$\text W_{ext} = \rho_w\text{Vgd} - \text{mgd}$

Substituting the values,

$\text W_{ext} = \left[1000 \times \dfrac{4}{3} \times \dfrac{22}{7}\left(\dfrac{3}{2} \times 10^{-2}\right)^3 - \dfrac{22}{7} \times 10^{-3}\right] \times 10 \times 0.7 \\[1em] = \dfrac{22}{7} \times 10^{-3}\left[\dfrac{9}{2} - 1\right] \times 10 \times 0.7$

$= 0.077\ \text J$

Hence option 1 is correct.

Option 2 : Neglecting the viscous force, the work done by gravity and by the buoyant force appears as the kinetic energy of the ball as it emerges,

$\text W_g + \text W_B = \text K_f - \text K_i = \dfrac{1}{2}\text{mv}^2$

Since this is numerically equal to the external work already calculated,

$\dfrac{1}{2} \times \dfrac{22}{7} \times 10^{-3} \times v^2 = 77 \times 10^{-3}$

$v^2 = 49 \quad \Rightarrow \quad v = 7\ \text{m s}^{-1}$

Hence option 2 is correct.

Option 3 : After emerging with speed v, the ball rises as a body thrown vertically upward,

$\text H = \dfrac{v^2}{2\text g} = \dfrac{49}{20} = 2.45\ \text m$

Since this is 2.45 m and not 1.4 m, option 3 is incorrect.

Option 4 : On working out the maximum viscous force from Stokes' law and comparing it with the net force (buoyant force − weight), the ratio does not come out to be $\dfrac{500}{9}$. Hence option 4 is incorrect.

A drop of liquid of radius R = 10^-2 m having surface tension $\text S = \dfrac{0.1}{4\pi}$ Nm^-1 divides itself into k identical drops. In this process the total change in the surface energy ΔU = 10^-3 J. If k = 10^α then the value of α is ............... .

Answer

Given,

• Radius of the drop, R = 10^-2 m
• Surface tension, $\text S = \dfrac{0.1}{4\pi}$ N m^-1
• Change in surface energy, ΔU = 10^-3 J
• Number of identical drops, k = 10^α

Let r be the radius of each small drop. Since the total volume remains unchanged,

$\dfrac{4}{3}\pi \text R^3 = \text k \times \dfrac{4}{3}\pi \text r^3$

$\text r^3 = \dfrac{\text R^3}{\text k} \quad \Rightarrow \quad \text r^2 = \dfrac{\text R^2}{\text k^{2/3}}$

The change in the surface energy is the surface tension multiplied by the increase in the total surface area,

$\Delta \text U = \text S(\text k \times 4\pi \text r^2) - \text S(4\pi \text R^2) = 4\pi \text S[\text k\text r^2 - \text R^2]$

Substituting the value of r²,

$\Delta \text U = 4\pi \text S\left[\text k \times \dfrac{\text R^2}{\text k^{2/3}} - \text R^2\right] = 4\pi \text S\text R^2[\text k^{1/3} - 1]$

Substituting the values,

$10^{-3} = 4\pi \times \dfrac{0.1}{4\pi} \times (10^{-2})^2 \times [\text k^{1/3} - 1] \\[1em] 10^{-3} = 0.1 \times 10^{-4} \times [\text k^{1/3} - 1]$

$\text k^{1/3} - 1 = \dfrac{10^{-3}}{10^{-5}} = 100$

$\text k^{1/3} = 101 \quad \Rightarrow \quad \text k = (101)^3 \simeq 10^{6}$

Comparing with k = 10^α,

Hence, the value of α is 6.

Two large, identical water tanks, 1 and 2, kept on the top of a building of height H, are filled with water up to height h in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are t₁ and t₂, respectively. If $\text H = \left(\dfrac{16}{9}\right)h$ then ratio $\dfrac{t_1}{t_2}$ is ............... .

Answer

Given,

• Height of water in each tank = h
• Height of the building = H, with $\text H = \dfrac{16}{9}\text h$
• Let A be the area of cross-section of each tank and a that of the hole

For tank 1 : The water flows out directly from the hole, so by Torricelli's theorem the velocity of efflux when the water level is at a height y is $v_1 = \sqrt{2\text{gy}}$.

By the equation of continuity,

$\text A\left(-\dfrac{\text{dy}}{\text{dt}}\right) = \text a\sqrt{2\text{gy}} \quad \Rightarrow \quad \text{dt} = -\dfrac{\text A}{\text a\sqrt{2\text g}} \times \dfrac{\text{dy}}{\sqrt{\text y}}$

Integrating from y = h to y = 0,

$\int_0^{\text t_1}\text{dt} = -\dfrac{\text A}{\text a\sqrt{2\text g}}\int_{\text h}^{0}\dfrac{\text{dy}}{\sqrt{\text y}} = \dfrac{\text A}{\text a\sqrt{2\text g}}\left[2\sqrt{\text y}\right]_0^{\text h}$

$\text t_1 = \dfrac{\text A}{\text a\sqrt{2\text g}} \times 2\sqrt{\text h} = \dfrac{\text A}{\text a}\sqrt{\dfrac{2\text h}{\text g}} \qquad \ldots(\text i)$

For tank 2 : The pipe carries the water down to the ground level, so the effective height driving the efflux is (y + H). Hence $v_2 = \sqrt{2\text g(\text y + \text H)}$, and

$\text A\left(-\dfrac{\text{dy}}{\text{dt}}\right) = \text a\sqrt{2\text g(\text y + \text H)}$

Integrating from y = h to y = 0,

$\text t_2 = \dfrac{\text A}{\text a\sqrt{2\text g}} \times 2\left[\sqrt{\text H + \text h} - \sqrt{\text H}\right]$

Substituting $\text H = \dfrac{16}{9}\text h$,

$\sqrt{\text H + \text h} = \sqrt{\dfrac{25\text h}{9}} = \dfrac{5}{3}\sqrt{\text h}, \qquad \sqrt{\text H} = \dfrac{4}{3}\sqrt{\text h}$

$\text t_2 = \dfrac{\text A}{\text a}\sqrt{\dfrac{2\text h}{\text g}}\left(\dfrac{5}{3} - \dfrac{4}{3}\right) = \dfrac{\text A}{\text a}\sqrt{\dfrac{2\text h}{\text g}} \times \dfrac{1}{3} \qquad \ldots(\text{ii})$

Dividing equation (i) by equation (ii),

$\dfrac{\text t_1}{\text t_2} = \dfrac{1}{\dfrac{1}{3}} = 3$

Hence, the ratio $\dfrac{\text t_1}{\text t_2}$ is 3.

A spherical soap bubble inside an air chamber at pressure P₀ = 10⁵ Pa has a certain radius so that the excess pressure inside the bubble is ΔP = 144 Pa. Now, the chamber pressure is reduced to $\dfrac{8\text P_0}{27}$ so that bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure ΔP in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP in Pa is ............... .

Answer

Given,

• Chamber pressure in the first case, P₀ = 10⁵ Pa
• Excess pressure in the first case, ΔP = 144 Pa
• Chamber pressure in the second case = $\dfrac{8\text P_0}{27}$
• The temperature remains unchanged

Let R and R₁ be the radii of the bubble in the two cases, and P, P₁ the corresponding pressures of the air inside it.

Case 1 :

$\text P - \text P_0 = \Delta \text P = \dfrac{4\text T}{\text R} \quad \Rightarrow \quad \text P = \text P_0 + \dfrac{4\text T}{\text R} \qquad \ldots(\text i)$

Case 2 :

$\text P_1 - \dfrac{8\text P_0}{27} = \Delta \text P_1 = \dfrac{4\text T}{\text R_1} \quad \Rightarrow \quad \text P_1 = \dfrac{4\text T}{\text R_1} + \dfrac{8\text P_0}{27} \qquad \ldots(\text{ii})$

Since the temperature remains unchanged, the process is isothermal and Boyle's law applies, P V = P₁V₁,

$\left(\text P_0 + \dfrac{4\text T}{\text R}\right)\dfrac{4}{3}\pi \text R^3 = \left(\dfrac{4\text T}{\text R_1} + \dfrac{8\text P_0}{27}\right)\dfrac{4}{3}\pi \text R_1^3 \qquad \ldots(\text{iii})$

Since the excess pressure is much smaller than the chamber pressure, the terms $\dfrac{4\text T}{\text R}$ and $\dfrac{4\text T}{\text R_1}$ may be neglected in comparison with the chamber pressures. Then equation (iii) becomes

$\text P_0\text R^3 = \dfrac{8\text P_0}{27}\text R_1^3$

$\dfrac{\text R_1^3}{\text R^3} = \dfrac{27}{8} \quad \Rightarrow \quad \text R_1 = \dfrac{3}{2}\text R$

The new excess pressure is therefore

$\Delta \text P_1 = \dfrac{4\text T}{\text R_1} = \dfrac{4\text T}{\dfrac{3}{2}\text R} = \dfrac{2}{3} \times \dfrac{4\text T}{\text R} = \dfrac{2}{3}\Delta \text P$

Substituting ΔP = 144 Pa,

$\Delta \text P_1 = \dfrac{2}{3} \times 144 \\[1em] = 96\ \text{Pa}$

Hence, the new excess pressure is 96 Pa.

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm² with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5 × 10³ kg/m³. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity = 10 m/s²)

Answer

Given,

• Area of the window, A = 100 cm² = 100 × 10^-4 m²
• Depth of the window, h = 3 m
• Density of water, ρ_w = 1 × 10³ kg m^-3
• Density of the liquid, ρ_l = 1.5 × 10³ kg m^-3
• g = 10 m s^-2

The pressure at the depth h on the liquid side is (P₀ + ρ_lgh) and that on the water side is (P₀ + ρ_wgh), where P₀ is the atmospheric pressure.

For the door to remain closed, the externally applied force together with the force due to water must balance the force due to the liquid,

$\text F_{ext} + \text F_w = \text F_l$

$\text F_{ext} = \text F_l - \text F_w = (\text P_0 + \rho_l\text{gh})\text A - (\text P_0 + \rho_w\text{gh})\text A$

The atmospheric pressure terms cancel out, so

$\text F_{ext} = (\rho_l - \rho_w)\text{ghA}$

Substituting the values,

$\text F_{ext} = (1500 - 1000) \times 10 \times 3 \times (100 \times 10^{-4}) \\[1em] = 500 \times 10 \times 3 \times 10^{-2}$

$= 150\ \text N$

Hence, a force of 150 N must be applied on the hinged door so that it does not get opened.

Statement (I): Viscosity of gases is greater than that of liquids.

Statement (II): Surface tension of a liquid decreases due to the presence of insoluble impurities.

In the light of the above statements, choose the most appropriate answer from the given options :

1. Statement I is correct but statement II is incorrect
2. Statement I is incorrect but Statement II is correct
3. Both Statement I and Statement II are incorrect
4. Both Statement I and Statement II are correct

Answer

Statement I is incorrect but Statement II is correct

Reason — Statement (I) is incorrect : Gases have a much smaller viscosity than liquids. At 20°C the coefficient of viscosity of water is 1.00 × 10^-2 poise, while that of air is only 1.81 × 10^-4 poise. This is why we can walk fast in air but not in water.

Statement (II) is correct : The surface tension of a liquid is affected by the presence of impurities. Sparingly soluble or insoluble impurities, such as detergent, oil or grease, decrease the surface tension of water. Only a highly soluble impurity, such as common salt, increases it.

Hence, Statement I is incorrect but Statement II is correct.

Statement (I): When speed of liquid is zero everywhere, pressure difference at any two points depends on equation P₁ − P₂ = ρg(h₁ − h₂).

Statement (II): In ventury tube shown 2gh = v₁² − v₂².

In the light of the above statements, choose the most appropriate answer from the given options.

1. Both Statement I and Statement II are correct.
2. Statement I is incorrect but Statement II is correct.
3. Both Statement I and Statement II are incorrect.
4. Statement I is correct but Statement II is incorrect.

Answer

Statement I is correct but Statement II is incorrect.

Reason — Applying Bernoulli's equation between the two points,

$\text P_1 + \rho \text{gh}_1 + \dfrac{1}{2}\rho v_1^2 = \text P_2 + \rho \text{gh}_2 + \dfrac{1}{2}\rho v_2^2$

where h₁ and h₂ are the heights of the points from any reference level.

Statement (I) is correct : When the speed of the liquid is zero everywhere, v₁ = v₂ = 0, and the above equation reduces to

$\text P_1 - \text P_2 = \rho \text g(\text h_1 - \text h_2)$

Statement (II) is incorrect : In the venturi tube shown, the two points lie at the same horizontal level, so h₁ = h₂ and Bernoulli's equation gives

$\text P_1 - \text P_2 = \dfrac{1}{2}\rho v_2^2 - \dfrac{1}{2}\rho v_1^2$

But the pressure difference is measured by the liquid column of height h, so P₁ − P₂ = ρgh. Therefore,

$\rho \text{gh} = \dfrac{1}{2}\rho v_2^2 - \dfrac{1}{2}\rho v_1^2$

$2\text{gh} = v_2^2 - v_1^2$

This is the opposite of what Statement II claims, since Statement II gives 2gh = v₁² − v₂².

Hence, Statement I is correct but Statement II is incorrect.