NEET Solid and Fluids MCQs & PYQs

Q: 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)

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

Q: 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)

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 $Q = Av = 2 \times 10^{-6} \times \sqrt{40} \approx 12.64 \times 10^{-6} \text{ m}^3/\text{s}$.

Q: 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)

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.

Q: If a soap bubble expands, the pressure inside the bubble : (2022)

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.

Q: 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)

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 $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.

Q: A liquid does not wet the solid surface if angle of contact is: (2020-Covid)

For a liquid to not wet a solid surface, it must form an obtuse angle of contact. Therefore, the angle of contact must be greater than $90^\circ$.

Q: A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5\text{ g}$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is: (2020)

Mass of water risen in capillary $m = \pi r^2 h \rho$. Since $h \propto \frac{1}{r}$, we get $m \propto r$. Thus, $m_2 = m_1 \left(\frac{r_2}{r_1}\right) = 5 \times \left(\frac{2r}{r}\right) = 10\text{ g}$.

Q: Water rises to height '$h$' in capillary tube. If the length of capillary tube above the surface of water is made less than '$h$', then: (2015 Re)

When a capillary tube is of insufficient length, the liquid rises to the top and changes its meniscus radius to maintain equilibrium ($hR = \text{constant}$). It does not overflow.

Practice NEET Solid and Fluids Questions

Question 11:

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)

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 12:

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)

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 13:

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)

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 14:

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)

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 15:

moderate

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

(2022)

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 16:

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)

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 17:

moderate

A liquid does not wet the solid surface if angle of contact is:

(2020-Covid)

For a liquid to not wet a solid surface, it must form an obtuse angle of contact. Therefore, the angle of contact must be greater than $90^\circ$.

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Question 18:

moderate

A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5\text{ g}$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:

(2020)

Mass of water risen in capillary $m = \pi r^2 h \rho$. Since $h \propto \frac{1}{r}$, we get $m \propto r$. Thus, $m_2 = m_1 \left(\frac{r_2}{r_1}\right) = 5 \times \left(\frac{2r}{r}\right) = 10\text{ g}$.

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Question 19:

moderate

Three liquids of densities $\rho_1$, $\rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $\theta_1$, $\theta_2$ and $\theta_3$ obey:

(2016 – II)

Capillary rise $h = \frac{2T \cos\theta}{r \rho g}$. Since $h, T, r, g$ are constant, $\cos\theta \propto \rho$. Since $\rho_1 > \rho_2 > \rho_3$, $\cos\theta_1 > \cos\theta_2 > \cos\theta_3$. For acute angles, this means $$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$$.

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Question 20:

moderate

Water rises to height ‘$h$’ in capillary tube. If the length of capillary tube above the surface of water is made less than ‘$h$’, then:

(2015 Re)

When a capillary tube is of insufficient length, the liquid rises to the top and changes its meniscus radius to maintain equilibrium ($hR = \text{constant}$). It does not overflow.

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