Q 11-09-138JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
The top of a water tank is open to air and its water level is maintained. It is giving out $0.74\ \text{m}^3$ water per minute through a circular opening of $2\ \text{cm}$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:
Answer: (B) $4.8\ \text{m}$
Flow rate $Q = \dfrac{0.74}{60} = 1.23\times10^{-2}\ \text{m}^3/\text{s}$ and area $A = \pi(0.02)^2 = 1.257\times10^{-3}\ \text{m}^2$.
$$v = \frac{Q}{A} = 9.81\ \text{m/s}$$
By Torricelli's law $v = \sqrt{2gh}$:
$$h = \frac{v^2}{2g} = \frac{96.3}{20} \approx 4.8\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics