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(C) The volume flow rate (or discharge) $Q$ from a hole at depth $y$ is given by $Q = A_v \cdot v$,where $A_v$ is the area of the hole and $v = \sqrt{

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$r(\pi)^{\frac{1}{2}}(3)^{\frac{1}{4}}$

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(C) The volume flow rate (or discharge) $Q$ from a hole at depth $y$ is given by $Q = A_v \cdot v$,where $A_v$ is the area of the hole and $v = \sqrt{2gy}$ is the velocity of efflux.For hole $A$ at depth $h$: Area $A_A = L^2$,velocity $v_A = \sqrt{2gh}$. So,$Q_A = L^2 \sqrt{2gh}$.For hole $B$ at depth $3h$: Area $A_B = \pi r^2$,velocity $v_B = \sqrt{2g(3h)} = \sqrt{6gh}$. So,$Q_B = \pi r^2 \sqrt{6gh}$.Given that the flow rate is the same,$Q_A = Q_B$:$L^2 \sqrt{2gh} = \pi r^2 \sqrt{6gh}$$L^2 = \pi r^2 \frac{\sqrt{6gh}}{\sqrt{2gh}} = \pi r^2 \sqrt{3}$Taking the square root of both sides:$L = \sqrt{\pi r^2 \sqrt{3}} = r \sqrt{\pi} (3)^{\frac{1}{4}} = r (\pi)^{\frac{1}{2}} (3)^{\frac{1}{4}}$.

$A$ vessel completely filled with water has holes $A$ and $B$ at depths $h$ and $3h$ from the top respectively. The 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 same. Then $L$ is equal to

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