NEET Fluid Dynamics MCQs & PYQs

Q: A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the holes are both same. Then, R is equal to:

According to Torricelli's Law, the velocity of efflux from a hole at a depth h is given by $$v = \sqrt{2gh}$$. Since the volume flow rate per second $ Q = \text{Area}\times\text{Velocity} $ is the same for both holes: $$A_{\text{square}} \cdot v_1 = A_{\text{circle}} \cdot v_2$$ $$L^2 \sqrt{2gy} = (\pi R^2) \sqrt{2g(4y)}$$ $$L^2 = \pi R^2 \cdot 2$$ $$R = \frac{L}{\sqrt{2\pi}}$$ Thus, the correct option is Option 1.

Practice NEET Fluid Dynamics Questions

Question 1:

moderate

A tank is filled with water to a height H. A hole is made in one of the wall at a depth D below the water surface. The distance x from the foot of the wall at which the stream of water strikes the ground is given by :

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Question 2:

moderate

A cylindrical tank has a hole of area 0.5 cm2 in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70 cm3/sec then the maximum height upto which water can rise in the tank is :

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Question 3:

moderate

A fully loaded Boeing aircraft has a mass of \[3.3\times 10^{5} kg\]. Its total wing area is 500 m2. It is in level flight with speed of 960 kmph. Estimate the pressure difference between the lower and upper surfaces of the wings :

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Question 4:

moderate

Figure shows a venturi meter, through which water is flowing. The speed of water at X is 2 cm/s. The speed of water at Y is :- (Take g = 1000 cm /s.s)

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Question 5:

moderate

A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of
water flowing out per second from the holes are both same. Then, R is equal to:

According to Torricelli's Law, the velocity of efflux from a hole at a depth $h$ is given by $v = \sqrt{2gh}$ .

Since the volume flow rate per second

\( $Q = \text{Area} \times \text{Velocity}$ is the same for both holes:

A_{\text{square}} \cdot v_1 = A_{\text{circle}} \cdot v_2

L^2 \sqrt{2gy} = (\pi R^2) \sqrt{2g(4y)}

L^2 = \pi R^2 \cdot 2

R = \frac{L}{\sqrt{2\pi}}

Thus, the correct option is Option 1.

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Question 6:

moderate

A tank is filled to a height H. The range of water coming out of a hole which is a depth H/4 from the surface of water level is :

The horizontal range of water emerging from a hole is given by the formula

$R = 2\sqrt{h(H - h)}$

R = 2\sqrt{\left(\frac{H}{4}\right)\left(\frac{3H}{4}\right)} = 2\left(\frac{\sqrt{3}H}{4}\right) = \frac{\sqrt{3}H}{2}

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Question 7:

moderate

A water tank resting on the floor has two small holes vertically one above the other. The holes are $h_1$ $text{cm}$ and $h_2$ $text{cm}$ above the floor. How high does water stand in the tank if the jets from the holes hits the floor at the same point ?

For equal horizontal range, the height $H$ of the water level in the tank must satisfy $h_1(H - h_1) = h_2(H - h_2)$. Solving for $H$ gives $H(h_2 - h_1) = h_2^2 - h_1^2$, which simplifies to $H = h_1 + h_2$.

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Question 8:

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 9:

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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