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

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Question

(A) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}

Vedclass · Accepted Answer

$\sqrt{2}-1$

Vedclass · Answer

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

List-$I$	List-$II$
$P$. Lift is accelerating vertically up.	$1$. $d = 1.2 \ m$
$Q$. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.	$2$. $d < 1.2 \ m$
$R$. Lift is moving vertically up with constant speed.	$3$. $d > 1.2 \ m$
$S$. Lift is falling freely.	$4$. No water leaks out of the jar.

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