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A square hole of side length $l$ is made at a depth of $h$ and a circular hole of radius $r$ is made at a depth of $4\,h$ from the surface of water in a water tank kept on a horizontal surface. If $l << h,\,r << h$ and the rate of water flow from the holes is the same, then $r$ is equal to
$\frac{l}{{\sqrt {2\pi } }}$
$\frac{l}{{\sqrt {3\pi } }}$
$\frac{l}{{{3\pi } }}$
$\frac{l}{{{2\pi } }}$
Solution
$As\,{A_1}{V_1} = {A_2}{V_2}\left( {principle\,of\,continuity} \right)$
$or,\,{\ell ^2}\sqrt {2gh} = \pi {r^2}\sqrt {2g \times 4h} $
$\left( {Efflux\,velocity = \sqrt {2gh} } \right)$
$\therefore \,{r^2} = \frac{{{\ell ^2}}}{{2\pi }}$ or $r = \sqrt {\frac{{{\ell ^2}}}{{2\pi }}} = \frac{\ell }{{\sqrt {2\pi } }}$
