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# A cylindrical vessel filled upto a height at the bottom of the vessel. Area of hole is $\bigg(\large\frac{1}{100}\bigg)$ that of the area of vessel is maintained at 2 m . What is the time required to empty water equal to volume of water initially present in the vessel.

$\begin {array} {1 1} (a)\;54\;seconds \\ (b)\;90\;seconds \\ (c)\;32 \;seconds \\ (d)\;15\;seconds \end {array}$

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## 1 Answer

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Now we know volume of water flowing out per second from the hole of area a is given by
Volume / second $= a v$
Where v= velocity of e flux
$v= \sqrt {2gh}$
$\therefore$ Time taken to empty water equal to volume of water initially present $t =\large\frac{A \times h}{av}$
$A$ = Area of vessel
$h$ = height of water
$t=\large\frac{A}{a} \sqrt{\frac {h}{2g}}$
$\quad= 100 \sqrt {\large\frac{2}{2 \times 9.8}}$
$\quad=\large\frac{100}{\sqrt {9.8}}$
$\quad \approx 32\;seconds$
answered Oct 8, 2013 by

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