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Physics · Fluid Mechanics

JEE Main 2025 — 3 April, Morning Shift — Question 57

Consider a completely full cylindrical water tank of height 1.6 m and of cross-sectional area 0.5 m20.5 \mathrm{~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/s2g=10 \mathrm{~m} / \mathrm{s}^{2} )

  1. Option A:

    3 m/s3 \mathrm{~m} / \mathrm{s}

  2. Option B:

    2 m/s2 \mathrm{~m} / \mathrm{s}

  3. Option C:

    5 m/s5 \mathrm{~m} / \mathrm{s}

  4. Option D:

    4 m/s4 \mathrm{~m} / \mathrm{s}

    Correct

Answer: D

Step-by-step solution

ΔP=12ρv2\Delta P=\frac{1}{2} \rho v^{2}

mgA+ρgΔh=12ρv2\frac{m g}{A}+\rho g \Delta h=\frac{1}{2} \rho v^{2}

5000.5+103×10×0.7=12×103v2\frac{500}{0.5}+10^{3} \times 10 \times 0.7=\frac{1}{2} \times 10^{3} v^{2}

1+7=12v21+7=\frac{1}{2} v^{2} v=4 m/sv=4 \mathrm{~m} / \mathrm{s}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Fluid Mechanics
Topic
Bernoulli's Equation and its Applications
Consider a completely full cylindrical water tank of height 1.6 m and… | JEE Main 2025 PYQ with Solution · DhiX AI