Question:

A large vessel completely filled with water has two holes ‘A’ and ‘B’ at depths \( h \) and \( 4h \) from the top. Hole ‘A’ is a square of side \( L \) and hole ‘B’ is a circle of radius \( R \). If from both the holes same quantity of water is flowing per second, then side of square hole is

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In fluid flow problems, discharge depends on both area and depth of hole.
Updated On: Jan 26, 2026
  • \( 2\pi R \)
  • \( \sqrt{2\pi R} \)
  • \( \sqrt{2\pi}\,R \)
  • \( \dfrac{R}{2} \)
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The Correct Option is C

Solution and Explanation

Step 1: Write Torricelli’s theorem.
Velocity of efflux at depth \( h \): \[ v = \sqrt{2gh} \]
Step 2: Write discharge formula.
\[ Q = A v \]
Step 3: Write discharge from both holes.
For hole A (depth \( h \), area \( L^2 \)): \[ Q_A = L^2 \sqrt{2gh} \]
For hole B (depth \( 4h \), area \( \pi R^2 \)): \[ Q_B = \pi R^2 \sqrt{2g(4h)} = \pi R^2 \cdot 2\sqrt{2gh} \]
Step 4: Equate discharges.
\[ L^2 \sqrt{2gh} = 2\pi R^2 \sqrt{2gh} \]
Step 5: Solve for \( L \).
\[ L^2 = 2\pi R^2 \Rightarrow L = \sqrt{2\pi}\,R \]
Step 6: Conclusion.
The side of square hole is \( \sqrt{2\pi}\,R \).
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