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(D) The volume flow rate $R_v$ is given by $R_v = A 	imes v$, where $A$ is the area of the hole and $v$ is the velocity of efflux given by Torricelli

Vedclass · Accepted Answer

$3.46$

Vedclass · Answer

(D) The volume flow rate $R_v$ is given by $R_v = A 	imes v$, where $A$ is the area of the hole and $v$ is the velocity of efflux given by Torricelli's law $v = \sqrt{2gh}$.For the square hole: Area $A_1 = x^2$, depth $h_1 = 2 \,m$. Velocity $v_1 = \sqrt{2g(2)} = 2\sqrt{g}$. Flow rate $R_{v1} = x^2 	imes 2\sqrt{g}$.For the triangular hole: Area $A_2 = \frac{\sqrt{3}}{4} 	imes (4)^2 = 4\sqrt{3} \,cm^2$, depth $h_2 = 6 \,m$. Velocity $v_2 = \sqrt{2g(6)} = \sqrt{12g} = 2\sqrt{3g}$. Flow rate $R_{v2} = 4\sqrt{3} 	imes 2\sqrt{3g} = 24\sqrt{g}$.Equating the flow rates: $x^2 	imes 2\sqrt{g} = 24\sqrt{g}$.$x^2 = 12$.$x = \sqrt{12} = 2\sqrt{3} \approx 3.46 \,cm$.

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