A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side 'L' at a depth '4y' from the top and the other is a circular hole of radius 'R' at a depth '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:

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$\frac{L}{\sqrt{2 π}}$

2 $πL$

$\sqrt{\frac{2}{π}} L$

$\frac{L}{2 π}$

answer is C.

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Detailed Solution

Let $v_{1} and v_{2}$ be the velocity of efflux from square and circular hole respectively. $S_{1} and S_{2}$ be cross-section areas of square and circular holes.

$v_{1} = \sqrt{8 gy} and v_{2} = \sqrt{2 g (y)}$

The volume of water coming out of square and circular hole per second is

$Q_{1} = v_{1} S_{1} = \sqrt{8 gy} L^{2};$

$Q_{2} = v_{2} S_{2} = \sqrt{2 g y} {πR}^{2}$

$Solve Q_{1} = Q_{2}$

$∴ R = \sqrt{\frac{2}{π}} . L$

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