Alarge tank is filled with water to a height $H$.A small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $H/ \eta , (\eta > 1)$ and it takes $T_2$ time to take out the rest of water. If $T_1 = T_2$ , then the value of $\eta$ is :
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$t=\frac{A}{a} \sqrt{\frac{2}{g}[\sqrt{H_{1}}-\sqrt{H_{2}}]}$
$\mathrm{Now}, T_{1}=\frac{A}{a} \sqrt{\frac{2}{g}[\sqrt{H}-\sqrt{\frac{H}{\eta}}]}$
and $T_{2}=\frac{A}{a} \sqrt{\frac{2}{g}}[\sqrt{\frac{H}{\eta}}-\sqrt{0}]$
According to problem $T_{1}=T_{2}$
$\therefore \operatorname{Sqrt}(H)-\sqrt{\frac{H}{\eta}}=\sqrt{\frac{H}{\eta}}-0 \Rightarrow \sqrt{H}=2 \sqrt{\frac{H}{\eta}} \Rightarrow \eta=4$
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