First slide

Fluid dynamics

Question

A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4 y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to

Moderate

A

$\frac{L}{\sqrt{(2 π)}}$
B

$2 πL$
C

L
D

$\frac{L}{2 π}$

Solution

We know that rate of flow of water = y A, where y is the speed of efflux and A is area of cross section of the hole, Equating the rate of flow, we have
$\sqrt{(2 gy)} \times L^{2} = \sqrt{(2 g \times 4 y)} {πR}^{2} L^{2} = 2 {πR}^{2} or R^{2} = L^{2} / 2 π or R = L / \sqrt{(2 π)}$

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