Fluid Dynamics

Practice Questions:

Question 1: 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 2: 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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