A large open tank has two holes in the wall. | vedclass

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(A) The volume flow rate (quantity of water per second) is given by $Q = A v$, where $A$ is the area of the hole and $v$ is the velocity of efflux.Acc

Vedclass · Accepted Answer

$\frac{L}{\sqrt{2 \pi}}$

Vedclass · Answer

(A) The volume flow rate (quantity of water per second) is given by $Q = A v$, where $A$ is the area of the hole and $v$ is the velocity of efflux.According to Torricelli's law, the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.For the square hole: Area $A_1 = L^2$ and depth $h_1 = y$. So, $v_1 = \sqrt{2gy}$.The flow rate $Q_1 = A_1 v_1 = L^2 \sqrt{2gy}$.For the circular hole: Area $A_2 = \pi R^2$ and depth $h_2 = 4y$. So, $v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$.The flow rate $Q_2 = A_2 v_2 = \pi R^2 (2\sqrt{2gy})$.Given that $Q_1 = Q_2$, we have:$L^2 \sqrt{2gy} = 2\pi R^2 \sqrt{2gy}$.Canceling $\sqrt{2gy}$ from both sides, we get $L^2 = 2\pi R^2$.Therefore, $R^2 = \frac{L^2}{2\pi}$, which gives $R = \frac{L}{\sqrt{2\pi}}$.

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