A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes T1 time to decrease the height of water to H/η (η>1) ; and it takes T2 time to take out the rest of water. If T1 = T2 , then the value of η is

The correct option is 4.

Let us derive the relation for the time taken to lower the height of water in a tank of uniform cross-sectional area $A$ through a small hole of area $a$ at the base. Let the instantaneous height of water in the tank be $h$.

According to Torricelli's Law, the velocity of efflux $v$ of water leaving the hole is given by:
$v=\sqrt{2gh}$

From the equation of continuity, the rate of decrease of water volume in the tank equals the rate of flow of water out of the hole:
$-A\frac{dh}{dt}=av$
Substituting the expression for $v$:
$-A\frac{dh}{dt}=a\sqrt{2gh}$

Separating the variables, we get:
$-h^{-1/2}dh=\frac{a}{A}\sqrt{2g}dt$

Integrating both sides from an initial height $h_1$ to a final height $h_2$ over a time interval $t$:
$-{\int }_{h_1}^{h_2}{h}^{-1/2}dh=\frac{a}{A}\sqrt{2g}{\int }_0^{t}dt$
${\left[-2\sqrt{h}\right]}_{h_1}^{h_2}=\frac{a}{A}\sqrt{2g}t$
$2\left(\sqrt{h_1}-\sqrt{h_2}\right)=\frac{a}{A}\sqrt{2g}t$

From this equation, we can see that the time $t$ required to lower the height from $h_1$ to $h_2$ is proportional to $\sqrt{h_1}-\sqrt{h_2}$:
$t\propto \left(\sqrt{h_1}-\sqrt{h_2}\right)$

For the first part of the process, the water level decreases from $H$ to $H/\eta$ in time $T_1$:
$T_1\propto \left(\sqrt{H}-\sqrt{\frac{H}{\eta }}\right)$

For the second part of the process, it takes time $T_2$ to empty the remaining water from height $H/\eta$ to $0$:
$T_2\propto \left(\sqrt{\frac{H}{\eta }}-0\right)$

Given that $T_1=T_2$, we can equate the proportional relations:
$\sqrt{H}-\sqrt{\frac{H}{\eta }}=\sqrt{\frac{H}{\eta }}$

Dividing both sides by $\sqrt{H}$ yields:
$1-\frac{1}{\sqrt{\eta }}=\frac{1}{\sqrt{\eta }}$

Rearranging the terms:
$1=\frac{2}{\sqrt{\eta }}$
$\sqrt{\eta }=2$

Squaring both sides:
$\eta =4$