Question

A vessel completely filled with water has holes A and B at depths \( h \) and \( 3h \) from the top, respectively. Hole A is a square of side \( L \), and B is a circle of radius \( r \). The water flowing out per second from both the holes is the same. Then \( L \) is equal to:

• A. \( L = 2^{3/4} \pi^{1/2} r \)
• B. \( L = 3^{1/4} \pi^{1/2} r \)
• C. \( L = 4^{1/2} \pi^{1/2} r \)
• D. \( L = 3^{1/2} \pi^{1/2} r \)

Hint:

Use Torricelli’s theorem \( v = \sqrt{2gh} \) to calculate efflux velocity and combine it with the area of the hole to equate flow rates.

Answer 1

The Correct Option is B

Step 1: Determine efflux velocities using Torricelli’s theorem.
The velocity of efflux \( v \) from a hole at depth \( h \) is given by \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity.For hole A at depth \( h \): \( v_A = \sqrt{2gh} \).For hole B at depth \( 3h \): \( v_B = \sqrt{2g(3h)} = \sqrt{6gh} \).Step 2: Relate flow rates.
Flow rate \( Q \) is calculated as \( Q = \text{Area} \times \text{Velocity} \).For hole A (square, side \( L \)): \( Q_A = L^2 \cdot v_A = L^2 \cdot \sqrt{2gh} \).For hole B (circle, radius \( r \)): \( Q_B = \pi r^2 \cdot v_B = \pi r^2 \cdot \sqrt{6gh} \).Given that \( Q_A = Q_B \): \( L^2 \cdot \sqrt{2gh} = \pi r^2 \cdot \sqrt{6gh} \).Step 3: Simplify the equation.
Cancel \( \sqrt{gh} \) from both sides: \( L^2 \cdot \sqrt{2} = \pi r^2 \cdot \sqrt{6} \).Divide by \( \sqrt{2} \): \( L^2 = \pi r^2 \cdot \sqrt{\frac{6}{2}} = \pi r^2 \cdot \sqrt{3} \).Take the square root of both sides: \( L = \sqrt{\pi r^2 \cdot \sqrt{3}} = r \cdot \pi^{1/2} \cdot (3^{1/4}) \).Therefore, \( L = 3^{1/4} \pi^{1/2} r \).