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A cylindrical vessel of cross-section A contains water to a height $\mathrm{h}$. There is a hole in the bottom of radius ' $a$ '. The time in which it will be emptied is:
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Verified Answer
The correct answer is:
$\frac{\sqrt{2} \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$
$\frac{\sqrt{2} \mathrm{~A}}{\pi \mathrm{a}^2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$
$$
\text { Let the rate of falling water level be }-\frac{d h}{d t}
$$
Initially at $t=0 ; h=h$
$$
t=t ; h=0
$$
Then, $A\left(-\frac{d h}{d t}\right)=\pi a^2 \cdot v$
$$
d t=-\frac{A}{\pi a^2 \sqrt{2 g h}} d h
$$
$[\because$ velocity of efflux of liquid $v=\sqrt{2 g h}]$ Integrating both sides
$$
\int_0^t d t=-\frac{A}{\sqrt{2 g} \pi a^2} \int_h^0 h^{-1 / 2} d h
$$
$$
[t]_0^t=-\frac{A}{\sqrt{2 g} \pi a^2} \cdot\left[\frac{h^{1 / 2}}{1 / 2}\right]_h^0
$$
$$
t=\frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}
$$
\text { Let the rate of falling water level be }-\frac{d h}{d t}
$$
Initially at $t=0 ; h=h$
$$
t=t ; h=0
$$
Then, $A\left(-\frac{d h}{d t}\right)=\pi a^2 \cdot v$
$$
d t=-\frac{A}{\pi a^2 \sqrt{2 g h}} d h
$$
$[\because$ velocity of efflux of liquid $v=\sqrt{2 g h}]$ Integrating both sides
$$
\int_0^t d t=-\frac{A}{\sqrt{2 g} \pi a^2} \int_h^0 h^{-1 / 2} d h
$$
$$
[t]_0^t=-\frac{A}{\sqrt{2 g} \pi a^2} \cdot\left[\frac{h^{1 / 2}}{1 / 2}\right]_h^0
$$
$$
t=\frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}
$$
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