A cylindrical vessel of cross-section A conta | vedclass

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(B) Let the rate of fall of the water level be $-\frac{dh}{dt}$.According to the equation of continuity,the volume of water leaving the hole per unit

Vedclass · Accepted Answer

$\frac{\sqrt{2}A}{\pi a^2}\sqrt{\frac{h}{g}}$

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(B) Let the rate of fall of the water level be $-\frac{dh}{dt}$.According to the equation of continuity,the volume of water leaving the hole per unit time is equal to the volume of water lost by the vessel per unit time.$A \left( -\frac{dh}{dt} \right) = a_{hole} \cdot v$Where $a_{hole} = \pi a^2$ is the area of the hole and $v = \sqrt{2gh}$ is the velocity of efflux (Torricelli's Law).So,$A \left( -\frac{dh}{dt} \right) = \pi a^2 \sqrt{2gh}$.Rearranging the terms to integrate:$dt = -\frac{A}{\pi a^2 \sqrt{2g}} h^{-1/2} dh$.Integrating from $t = 0$ to $T$ (total time) and $h = h$ to $0$:$\int_0^T dt = -\frac{A}{\pi a^2 \sqrt{2g}} \int_h^0 h^{-1/2} dh$.$T = -\frac{A}{\pi a^2 \sqrt{2g}} \left[ \frac{h^{1/2}}{1/2} \right]_h^0$.$T = -\frac{A}{\pi a^2 \sqrt{2g}} \cdot 2 [0 - \sqrt{h}] = \frac{2A}{\pi a^2 \sqrt{2g}} \sqrt{h}$.$T = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}} = \frac{\sqrt{2}A}{\pi a^2} \sqrt{\frac{h}{g}}$.

$A$ cylindrical vessel of cross-section $A$ contains water to a height $h$. There is a hole in the bottom of radius $a$. The time in which it will be emptied is

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