Water is filled in a container upto height of $3\,m$. A small hole of area $‘A_0’$ is punched in the wall of the container at a height $52.5\, cm$ from the bottom. The cross sectional area of the container is $A$. If $A_0/A = 0.1$ then $v^2$ is......... $m^2/s^2$ (where $v$ is the velocity of water coming out of the hole)
AIIMS 2008,AIIMS 2016, Medium
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The square of the velocity of flux
${v^2} = \frac{{2gh}}{{1 - {{\left( {\frac{{{A_0}}}{A}} \right)}^2}}}$
$ = \frac{{2 \times 10 \times 2.475}}{{1 - {{\left( {0.1} \right)}^2}}} = 50{m^2}/{s^2}$
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