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

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(C) The volume flow rate (quantity of water per second) is given by the product of the area of the hole $(A)$ and the velocity of efflux $(v)$.Accordi

Vedclass · Accepted Answer

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

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(C) The volume flow rate (quantity of water per second) is given by the product of the area of the hole $(A)$ and the velocity of efflux $(v)$.According to Torricelli's law,the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.For the square hole: Area $A_1 = a^2$,depth $h_1 = x$. Velocity $v_1 = \sqrt{2gx}$.Flow rate $Q_1 = A_1 v_1 = a^2 \sqrt{2gx}$.For the circular hole: Area $A_2 = \pi r^2$,depth $h_2 = 4x$. Velocity $v_2 = \sqrt{2g(4x)} = 2\sqrt{2gx}$.Flow rate $Q_2 = A_2 v_2 = \pi r^2 (2\sqrt{2gx})$.Given $Q_1 = Q_2$,we have:$a^2 \sqrt{2gx} = 2\pi r^2 \sqrt{2gx}$.Dividing both sides by $\sqrt{2gx}$:$a^2 = 2\pi r^2$.$r^2 = \frac{a^2}{2\pi}$.$r = \frac{a}{\sqrt{2\pi}}$.

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