JEE Advanced 2024 — Paper 1 — PHY-III solution

JEE Advanced2024Paper 1PHY-III

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 the tanks $1$ and $2$ through the holes, the times taken to empty the tanks are $t_{1}$ and $t_{2}$ , respectively. If $H = (\frac{16}{9}) h$ , then the ratio $t_{1} / t_{2}$ is _______.

Your answer

Solution

For a tank with cross-section $A$ and hole area $a$ , the standard drain-time formula is $T = \frac{2 A}{a 2 g} (h_{i} - h_{f})$ , where $h_{i}$ and $h_{f}$ are the initial and final effective heads above the exit.

Tank 1: Effective head is just the water column. $h_{i} = h$ , $h_{f} = 0$ :

$t_{1} = \frac{2 A}{a 2 g} h$ .

Tank 2: The exit is at the ground via the pipe, so the effective head equals water level above ground. Initially $h_{i} = h + H = h + \frac{16 h}{9} = \frac{25 h}{9}$ ; the tank empties when the water level inside reaches the pipe inlet, which is at height $H$ above the ground, so $h_{f} = H = \frac{16 h}{9}$ .

$t_{2} = \frac{2 A}{a 2 g} (\frac{25 h}{9} - \frac{16 h}{9}) = \frac{2 A h}{a 2 g} \cdot \frac{5 - 4}{3} = \frac{2 A h}{3 a 2 g}$ .

Therefore $\frac{t _{1}}{t _{2}} = 3$ .

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →