A cylindrical tank of height $0.4\,m$ is open at the top and has a diameter $0.16\,m$ . Water is filled in it up to a height of $0.16\,m$ . How long it will take to empty the tank through a hole of radius $5 \times 10^{-3}\,m$ in its bottom .......... $\sec$
Diffcult
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$A_{1} v_{1}=A_{2} v_{2}$

$\pi \mathrm{R}^{2} \mathrm{dh} / \mathrm{dt}=\pi \mathrm{r}^{2} \mathrm{v}$        $...(i)$

$\mathrm{v}=\sqrt{2 \mathrm{gh}}$       $...(ii)$

from equation $(ii)$ put the value of $v$ in equation $( i)$

$-\pi \mathrm{R}^{2} \mathrm{dh} / \mathrm{dt}=\pi \mathrm{r}^{2} \sqrt{2 \mathrm{gh}}$

$\Rightarrow-\int \frac{\mathrm{R}^{2} \mathrm{dh}}{\mathrm{r}^{2} \sqrt{2 \mathrm{gh}}}=\int \mathrm{dt} \Rightarrow-\frac{\mathrm{R}^{2}}{\mathrm{r}^{2} \sqrt{2 \mathrm{g}}} \int_{\mathrm{h}}^{0} \frac{\mathrm{dh}}{\sqrt{\mathrm{h}}}=\int_{0}^{\mathrm{t}} \mathrm{dt}$

on solving $t=46.26$ second

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