Solid and Fluids

Practice Questions:

Question 111: moderate

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?

(2018)

1. $4 F$

2. $6 F$

3. $9 F$

4. $F$

View Answer

Volume $V = A_1 L_1 = A_2 L_2 \Rightarrow A L_1 = 3A L_2 \Rightarrow L_2 = L_1/3$. Force $F = \frac{Y A \Delta l}{L_1}$. For the second wire, $F' = \frac{Y (3A) \Delta l}{L_1/3} = 9 \left( \frac{Y A \Delta l}{L_1} \right) = 9F$.

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Question 112: easy

The Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:

(2015 Re)

1. $1 : 1$

2. $1 : 2$

3. $2 : 1$

4. $4 : 1$

View Answer

We know $\Delta L = \frac{FL}{AY}$. Since $L$, $A$, and $\Delta L$ are the same for both wires, $F \propto Y$. Therefore, $$\frac{F_s}{F_b} = \frac{Y_s}{Y_b} = \frac{2}{1} = 2:1$$.

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Question 113: easy

Copper of fixed volume $V$ is drawn into wire of length $l$. When this wire is subjected to a constant force $F$, the extension produced in the wire is $\Delta l$. Which of the following graphs is a straight line?

(2014)

1. $\Delta l \text{ versus } 1/l$

2. $\Delta l \text{ versus } l^2$

3. $\Delta l \text{ versus } 1/l^2$

4. $\Delta l \text{ versus } l$

View Answer

Extension $\Delta l = \frac{Fl}{AY}$. Since volume $V = Al$, we have $A = V/l$. Substituting this, $$\Delta l = \frac{Fl}{(V/l)Y} = \frac{Fl^2}{VY}$$. Thus, $\Delta l \propto l^2$, meaning the graph of $\Delta l$ versus $l^2$ is a straight line.

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Question 114: moderate

A barometer is constructed using a liquid (density = $760 \text{ kg/m}^3$). What would be the height of the liquid column, when a mercury barometer reads $76 \text{ cm}$? (density of mercury = $13600 \text{ kg/m}^3$)

(2020-Covid)

1. $13.6 \text{ m}$

2. $136 \text{ m}$

3. $0.76 \text{ m}$

4. $1.36 \text{ m}$

View Answer

Equating pressures, we get $h_1 \rho_1 g = h_2 \rho_2 g$. Substituting the given values, $$h_1 \times 760 = 0.76 \times 13600$$. Solving for $h_1$, we obtain $h_1 = 13.6 \text{ m}$.

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Question 115: moderate

Two non-mixing liquids of densities $\rho$ and $n\rho$ ($n > 1$) are put in a container. The height of each liquid is $h$. A solid cylinder of length $L$ and density $d$ is put in this container. The cylinder floats with its axis vertical and length $pL$ ($p < 1$) in the denser liquid. The density $d$ is equal to:

(2016 – I)

1. $\{1 + (n + 1)p\}\rho$

2. $\{2 + (n + 1)p\}\rho$

3. $\{2 + (n - 1)p\}\rho$

4. $\{1 + (n - 1)p\}\rho$

View Answer

In equilibrium, the weight of the cylinder is balanced by the buoyant force from both liquids. $d \cdot A \cdot L \cdot g = \rho \cdot A \cdot (L - pL) \cdot g + n\rho \cdot A \cdot pL \cdot g$. Dividing by $A \cdot L \cdot g$, we get $d = \rho (1 - p) + n\rho p = \{1 + (n - 1)p\}\rho$.

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Question 116: moderate

A small hole of area of cross-section $2 \text{ mm}^2$ is present near the bottom of a fully filled open tank of height $2 \text{ m}$. Taking $g = 10 \text{ m/s}^2$, the rate of flow of water through the open hole would be nearly

(2019)

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

2. $8.9 \times 10^{-6} \text{ m}^3/\text{s}$

3. $2.23 \times 10^{-6} \text{ m}^3/\text{s}$

4. $6.4 \times 10^{-6} \text{ m}^3/\text{s}$

View Answer

Velocity of efflux is $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 2} = \sqrt{40} \text{ m/s}$. The rate of flow is given by $Q = Av = 2 \times 10^{-6} \times \sqrt{40} \approx 12.64 \times 10^{-6} \text{ m}^3/\text{s}$.

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Question 117: moderate

A wind with speed $40 \text{ m/s}$ blows parallel to the roof of a house. The area of the roof is $250 \text{ m}^2$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be ($P_{air} = 1.2 \text{ kg/m}^3$):

(2015)

1. $4.8 \times 10^5 \text{ N}$, upwards

2. $2.4 \times 10^5 \text{ N}$, upwards

3. $2.4 \times 10^5 \text{ N}$, downwards

4. $4.8 \times 10^5 \text{ N}$, downwards

View Answer

By Bernoulli's theorem, the pressure difference is $\Delta P = \frac{1}{2}\rho v^2 = \frac{1}{2} \times 1.2 \times (40)^2 = 960 \text{ N/m}^2$. The upward force is $F = \Delta P \times A = 960 \times 250 = 2.4 \times 10^5 \text{ N}$ directed upwards.

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Question 118: moderate

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

(2022)

1. Is equal to the atmospheric pressure

2. Decreases

3. Increases

4. Remains the same

View Answer

The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{R}$. As the bubble expands, its radius $R$ increases. Therefore, the excess pressure (and total pressure inside) decreases.

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Question 119: moderate

A soap bubble, having radius of $1 \text{ mm}$, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2} \text{ N/m}$. The pressure inside the bubble equals at a point $Z_0$ below the free surface of water in a container. Taking $g = 10 \text{ m/s}^2$, density of water $= 10^3 \text{ kg/m}^3$, the value of $Z_0$ is :

(2019)

1. $100 \text{ cm}$

2. $10 \text{ cm}$

3. $1 \text{ cm}$

4. $0.5 \text{ cm}$

View Answer

Pressure inside the bubble is $P = P_0 + \frac{4T}{r}$. Pressure at depth $Z_0$ is $P = P_0 + \rho g Z_0$. Equating them, $\rho g Z_0 = \frac{4T}{r}$. Solving gives $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.

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Question 120: moderate

A rectangular film of liquid is extended from $(4 \text{ cm} \times 2 \text{ cm})$ to $(5 \text{ cm} \times 4 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$, the value of the surface tension of the liquid is:

(2016 – II)

1. $0.2 \text{ Nm}^{-1}$

2. $8.0 \text{ Nm}^{-1}$

3. $0.250 \text{ Nm}^{-1}$

4. $0.125 \text{ Nm}^{-1}$

View Answer

Work done in stretching a liquid film is $W = T \times 2\Delta A$ (since it has two surfaces). The change in area is $\Delta A = (5 \times 4) - (4 \times 2) = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$. Thus, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.

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