JEE Main 2025 Apr 3 Shift 1, Physics Q2: Hydrodynamic

JEE Main2025Apr 3, Shift 1Physics

Consider a completely full cylindrical water tank of height 1.6 m and 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 $^{2}$ )

A
3 m/s
B
5 m/s
C
2 m/s
D
4 m/s

Solution

Apply Bernoulli's equation between the top free surface (point 1) and the hole (point 2). The water column above the hole is $h = 1.6 - 0.9 = 0.7$ m.

The 50 kg load adds an extra pressure on top: $Δ p = \frac{M g}{A} = \frac{50 \times 10}{0.5} = 1000$ Pa.

With $v_{1} \approx 0$ (large tank cross-section) and both ends at atmospheric pressure outside,

$\frac{M g}{A} + ρ g h = \frac{1}{2} ρ v_{2}^{2}$

$1000 + 1000 \times 10 \times 0.7 = \frac{1}{2} (1000) v_{2}^{2}$

$1000 + 7000 = 500 v_{2}^{2} \Rightarrow v_{2}^{2} = 16 \Rightarrow v_{2} = 4$ m/s.

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JEE Main 2025 Apr 3 Shift 1, Physics Q2: Hydrodynamic

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