A large tank filled with water to a height h | vedclass

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(C) According to Torricelli's law, the velocity of efflux is $v = \sqrt{2gh}$.Let $A$ be the area of the tank and $a$ be the area of the hole. The rat

Vedclass · Accepted Answer

$\sqrt{2}-1$

Vedclass · Answer

(C) According to Torricelli's law, the velocity of efflux is $v = \sqrt{2gh}$.Let $A$ be the area of the tank and $a$ be the area of the hole. The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$.Rearranging the terms, $dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.Integrating from height $h_1$ to $h_2$, the time taken is $t = \int_{h_2}^{h_1} \frac{A}{a \sqrt{2g}} h^{-1/2} dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{h_2}^{h_1} = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h_1} - \sqrt{h_2})$.For the first interval ($h$ to $h/2$): $t_1 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{h/2}) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1 - 1/\sqrt{2})$.For the second interval ($h/2$ to $0$): $t_2 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h/2} - 0) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1/\sqrt{2})$.Taking the ratio: $t_1/t_2 = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \frac{(\sqrt{2}-1)/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}-1$.

$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $h$ to $h/2$ and from $h/2$ to $0$ is

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