Two large, identical water tanks, 1 and 2, kept on the top of a building of height 𝐻, are filled with water up to height ℎ in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are 𝑡₁ and 𝑡₂, respectively. If 𝐻 = (16/9 ) ℎ, then the ratio 𝑡₁/𝑡₂ is _____.

The correct answer is 3.

Step-by-step Derivation:

Let the cross-sectional area of both identical tanks be $A$ and the area of the hole (as well as the internal area of the pipe) be $a$.

Case 1: Tank 1 (Without Pipe)

For Tank 1, the water level decreases from height $h$ to $0$. At any instantaneous water level height $y$ from the bottom of the tank, the velocity of efflux $v_1$ through the hole is given by Torricelli's law:
$v_1=\sqrt{2gy}$
The rate of flow of water from the tank is:
$-A\frac{dy}{dt}=a{v}_1=a\sqrt{2gy}$
Rearranging and integrating from $y=h$ (at $t=0$) to $y=0$ (at $t=t_1$):
$-{\int }_h^0{y}^{-1/2}dy=\frac{a}{A}\sqrt{2g}{\int }_0^{t_1}dt$
$2\sqrt{h}=\frac{a}{A}\sqrt{2g}{t}_1$
Thus, the time taken to empty Tank 1 is:
$t_1=\frac{A}{a}\sqrt{\frac{2h}{g}}$

Case 2: Tank 2 (With Pipe ending at Ground Level)

For Tank 2, a pipe of cross-sectional area $a$ is connected to the hole at the bottom (which is at height $H$ above the ground) and extends to the ground level (height $0$).
At any instant when the height of the water in the tank is $y$ from the bottom of the tank, the total height of the water column relative to the ground is $y+H$.
Applying Bernoulli's equation between the top surface of the water in the tank and the exit of the pipe at ground level:
$P_{\text{atm}}+\rho g(y+H)+0=P_{\text{atm}}+\frac{1}{2}\rho v_2^2$
This gives the velocity of efflux at the ground level exit of the pipe:
$v_2=\sqrt{2g(y+H)}$
The rate of decrease of water in the tank is:
$-A\frac{dy}{dt}=a{v}_2=a\sqrt{2g(y+H)}$
Rearranging and integrating from $y=h$ (at $t=0$) to $y=0$ (at $t=t_2$):
$-{\int }_h^0\frac{dy}{\sqrt{y+H}}$=aA2g∫0t2dt
$2\left[\sqrt{h+H}-\sqrt{H}\right]=\frac{a}{A}\sqrt{2g}{t}_2$
Thus, the time taken to empty Tank 2 is:
$t_2=\frac{A}{a}\sqrt{\frac{2}{g}}\left(\sqrt{h+H}-\sqrt{H}\right)$

Finding the Ratio of Times:

Dividing the expression for $t_1$ by $t_2$:
$\frac{t_1}{t_2}=\frac{\sqrt{h}}{\sqrt{h+H}-\sqrt{H}}$
Given that $H=\frac{16}{9}h$, we compute the term in the denominator:
$\sqrt{H}=\sqrt{\frac{16}{9}h}=\frac{4}{3}\sqrt{h}$
$\sqrt{h+H}=\sqrt{h+\frac{16}{9}h}=\sqrt{\frac{25}{9}h}=\frac{5}{3}\sqrt{h}$
Substituting these values back into the ratio:
$\frac{t_1}{t_2}=\frac{\sqrt{h}}{\frac{5}{3}\sqrt{h}-\frac{4}{3}\sqrt{h}}=\frac{1}{\frac{5}{3}-\frac{4}{3}}=\frac{1}{1/3}=3$