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 $h_1 = 13.6 \text{ m}$.
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 $d = \rho (1 - p) + n\rho p = \{1 + (n - 1)p\}\rho$.
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 $Q = Av = 2 \times 10^{-6} \times \sqrt{40} \approx 12.64 \times 10^{-6} \text{ m}^3/\text{s}$.
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.
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.
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 $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.
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, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.
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)
1. $\frac{\pi}{2} < \theta_1 < \theta_2 < \theta_3 < \pi$
2. $\pi > \theta_1 > \theta_2 > \theta_3 > \frac{\pi}{2}$
3. $\frac{\pi}{2} > \theta_1 > \theta_2 > \theta_3 \ge 0$
4. $0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$
View Answer
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 $$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$$.