A large 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 $\frac{H}{\eta }\,(\eta > 1)$; and it takes ${T_2}$ time to take out the rest of water. If ${T_1} = {T_2}$, then the value of $\eta $ is
A$2$
B$3$
C$4$
D$2\sqrt 2 $
Diffcult
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C$4$
c (c) $t = \frac{A}{a}\sqrt {\frac{2}{g}} \left[ {\sqrt {{H_1}} - \sqrt {{H_2}} } \right]$
Now, ${T_1} = \frac{A}{a}\sqrt {\frac{2}{g}} \left[ {\sqrt H - \sqrt {\frac{H}{\eta }} } \right]$
and ${T_2} = \frac{A}{a}\sqrt {\frac{2}{g}} \left[ {\sqrt {\frac{H}{\eta }} - \sqrt 0 } \right]$
According to problem ${T_1} = {T_2}$
$\sqrt H - \sqrt {\frac{H}{\eta }} = \sqrt {\frac{H}{\eta }} - 0$==> $\sqrt H = 2\sqrt {\frac{H}{\eta }} \Rightarrow \eta = 4$
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