JEE Advanced 2024 Paper 1 · Q12 · Bernoulli & Continuity

Question

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 $t_1$ and $t_2$, respectively. If $H = \left(\dfrac{16}{9}\right)h$, then the ratio $t_1/t_2$ is _____.

SolveKar AI · Accepted Answer

Tank 1 (no pipe): efflux speed at level $y$ above the hole is $\sqrt{2gy}$ (Torricelli). Mass-conservation $-A\dfrac{dy}{dt} = a\sqrt{2gy}$. Integrating from $y = h$ to $0$: $t_1 = \dfrac{A}{a}\sqrt{\dfrac{2h}{g}} = \dfrac{2A\sqrt{h}}{a\sqrt{2g}}$.

Tank 2 (with pipe to ground level $H$ below the hole): Bernoulli between water surface and pipe outlet at the ground gives efflux speed $\sqrt{2g(y + H)}$. So $-A\dfrac{dy}{dt} = a\sqrt{2g(y+H)}$. Integrating from $h$ to $0$: $t_2 = \dfrac{2A}{a\sqrt{2g}}\bigl[\sqrt{h+H} - \sqrt{H}\bigr]$.

With $H = (16/9)h$: $\sqrt{h+H} = \sqrt{h + 16h/9} = \sqrt{25h/9} = (5/3)\sqrt h$ and $\sqrt H = (4/3)\sqrt h$. So $t_2 = \dfrac{2A}{a\sqrt{2g}}\cdot \dfrac{1}{3}\sqrt h$. Therefore $\dfrac{t_1}{t_2} = \dfrac{\sqrt h}{(1/3)\sqrt h} = 3$.

JEE Advanced 2024 Paper 1 · Q12 · Bernoulli & Continuity

Solution