Question

A large open tank containing water has two holes to its wall. A square hole of side \( a \) is made at a depth \( y \) and a circular hole of radius \( r \) is made at a depth \( 16y \) from the surface of water. If equal amount of water comes out through both the holes per second, then the relation between \( r \) and \( a \) will be

• A. \( r = \frac{2a}{\pi} \)
• B. \( r = \frac{a}{\sqrt{\pi}} \)
• C. \( r = \frac{a}{2\pi} \)
• D. \( r = \frac{2a}{\sqrt{\pi}} \)

Hint:

The flow rate through an orifice depends on the area of the hole and the height of the liquid above it. Use Torricelli's law for flow calculations.

Answer 1

The Correct Option is D

Step 1: Flow rate from holes.
The flow rate \( Q \) through an orifice is given by Torricelli's law: \[ Q = C_d A \sqrt{2 g h} \] where \( C_d \) is the discharge coefficient, \( A \) is the area of the hole, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid above the hole.
Step 2: Apply to both holes.
For the square hole, the area is \( A_1 = a^2 \), and the depth is \( y \). For the circular hole, the area is \( A_2 = \pi r^2 \), and the depth is \( 16y \). The flow rates should be equal, so: \[ a^2 \sqrt{2 g y} = \pi r^2 \sqrt{2 g (16y)} \] Simplifying: \[ a^2 = 4 \pi r^2 \] Thus: \[ r = \frac{a}{2\sqrt{\pi}} \]
Step 3: Conclusion.
The correct relation between \( r \) and \( a \) is \( r = \frac{2a}{\sqrt{\pi}} \), so the correct answer is (D).