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As square box of water has a small hole located in one of the bottom corner. When the box is full and sitting on level surface complete opening of the hole results in a flow of water with speed $v_0.$ When box is half empty, it is lilted by $45^{\circ}$. So that the hole is at lowest position. Now the water will flow with speed of

\[\begin {array} {1 1} (a)\;v_0 \\ (b)\;\frac{v_0}{2} \\ (c)\;\frac{v_0}{\sqrt 2}\\ (d)\;\frac{v_0}{4 \sqrt 2} \end {array}\]

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The velocity of flow depends on the depth of water above the hole.
$v_0=\sqrt {2gh}$
$v=\sqrt {2g \frac{h}{\sqrt 2}}$
$\quad= \large\frac{v_0}{4 \sqrt 2}$
Hence d is the correct answer.


answered Oct 3, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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