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A square hole of side length $l$ is made at a depth of $y$ and a circular hole s made at a depth of $4 y$ from the surface of water tank, kept on a horizontal surface. If equal amount of water comes out of the vessel through the holes per second. Then the radius of circular hole is equal to $(r, l < < y)$ :

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$\frac{l}{\sqrt{π}}$

$\frac{l}{\sqrt{2 π}}$

$\frac{l}{2}$

$\frac{l}{\sqrt{2}}$

answer is D.

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Detailed Solution

Equation of continuity at 1 and 2,

$r_{1} = A_{1} v_{1}$

and $r_{2} = A_{2} v_{2}$

it is given that

$r_{1} = r_{2}$

( $r =$ rate of flow of liquid through the holes).

$∴$ $A_{1} v_{1} = A_{2} v_{2}$

Where the velocities of efflux are,

$v_{1} = \sqrt{2 g y}$ and $v_{2} = \sqrt{2 g (4 y)}$

And areas of cross-section of the holes are,

$A_{1} = l^{2}$ and $A_{2} = π R^{2}$

Putting all the values, we obtain,

$l^{2} \sqrt{(2 g y)} = {\sqrt{2 g (4 y)}} π R^{2}$

$∴$ $R = l / \sqrt{2 π}$

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