Physics Q32 – JEE-MAIN 2025

Home Physics Mechanical Properties of FluidsQuestion #271

Consider a completely full cylindrical water tank of height

1.6 m

and of cross-sectional area

0.5 m^{2}

. It has a small hole in its side at a height

90 cm

from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load

50 kg

is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is: (

g = 10 {m/s}^{2}

)

Clue 1 of 3: Apply Bernoulli's Principle

Consider two points: the free surface of the water inside the tank and the exit point of the hole.

🥷

Ninja Strategy•Lower Bound Estimation

First, calculate the efflux velocity using Torricelli's law (without the applied load) to establish a lower bound. The actual velocity with the load must be greater than this value, which helps eliminate options.

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Step 1: Identify Parameters and Apply Bernoulli's Principle✦ Active

Given tank height $H = 1.6 m$ , hole height $h_{h o l e} = 90 cm = 0.9 m$ , tank area $A = 0.5 m^{2}$ , load mass $m_{l o a d} = 50 kg$ , and $g = 10 {m/s}^{2}$ . The density of water is $ρ = 1000 {kg/m}^{3}$ . We apply Bernoulli's principle between the water surface (point 1) and the hole (point 2). Since the hole area is negligible, the velocity of the water surface $v_{1} \approx 0$ . The pressure at the hole $P_{2}$ is atmospheric pressure $P_{a t m}$ . The pressure at the surface $P_{1}$ is $P_{a t m} + P_{l o a d}$ , where $P_{l o a d} = \frac{m_{l o a d} g}{A}$ .

P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}

Step 2: Substitute Values and Simplify○ Expand

Substitute the identified pressures, velocities, and heights into Bernoulli's equation. Let the bottom of the tank be the reference for height, so $h_{1} = H$ and $h_{2} = h_{h o l e}$ .

(P_{a t m} + \frac{m_{l o a d} g}{A}) + \frac{1}{2} ρ (0)^{2} + ρ g H = P_{a t m} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{h o l e}

Cancel $P_{a t m}$ from both sides and rearrange to solve for $v_{2}^{2}$ :

\frac{m_{l o a d} g}{A} + ρ g H = \frac{1}{2} ρ v_{2}^{2} + ρ g h_{h o l e}

\frac{1}{2} ρ v_{2}^{2} = \frac{m_{l o a d} g}{A} + ρ g (H - h_{h o l e})

💡 Teacher's Secret Hint

Remember to use consistent units (SI units) for all quantities.

Step 3: Calculate the Efflux Velocity○ Expand

Now, substitute the numerical values into the equation:

\frac{1}{2} (1000) v_{2}^{2} = \frac{50 \times 10}{0.5} + (1000) \times 10 \times (1.6 - 0.9)

500 v_{2}^{2} = \frac{500}{0.5} + 10000 \times 0.7

500 v_{2}^{2} = 1000 + 7000

500 v_{2}^{2} = 8000

v_{2}^{2} = \frac{8000}{500} = 16

v_{2} = \sqrt{16} = 4 m/s

💡 Teacher's Secret Hint

Double-check your arithmetic, especially with unit conversions and division by decimals.

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Physics Question 32 – JEE-MAIN 2025

Clue 1 of 3: Apply Bernoulli's Principle

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