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(C) The time $t$ taken to change the water level from $H_1$ to $H_2$ in a tank with cross-sectional area $A$ and orifice area $a$ is given by $t = \fr

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$4$

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(C) The time $t$ taken to change the water level from $H_1$ to $H_2$ in a tank with cross-sectional area $A$ and orifice area $a$ is given by $t = \frac{A}{a}\sqrt{\frac{2}{g}} [\sqrt{H_1} - \sqrt{H_2}]$.For the first interval,the height changes from $H$ to $H/\eta$. Thus,$T_1 = \frac{A}{a}\sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{H/\eta}]$.For the second interval,the height changes from $H/\eta$ to $0$. Thus,$T_2 = \frac{A}{a}\sqrt{\frac{2}{g}} [\sqrt{H/\eta} - 0]$.Given $T_1 = T_2$,we equate the expressions:$\sqrt{H} - \sqrt{H/\eta} = \sqrt{H/\eta} - 0$.$\sqrt{H} = 2\sqrt{H/\eta}$.Squaring both sides,$H = 4(H/\eta)$.Therefore,$\eta = 4$.

$A$ 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 $H/\eta$ (where $\eta > 1$); and it takes $T_2$ time to drain the rest of the water. If $T_1 = T_2$,then the value of $\eta$ is:

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