Consider a water tank as shown in the figure. It's cross-sectional area is $0.4\ \text{m}^2$. The tank has an opening $B$ near the bottom whose cross-section area is $1\ \text{cm}^2$. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening $B$ is $v\ \text{m s}^{-1}$. The value of $v$, to the nearest integer, is ______. [Take the value of $g$ to be $10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 3
Extra pressure from the load: $\dfrac{24\times10}{0.4} = 600$ Pa. The tank is much wider than the opening, so the speed of the top surface is negligible. By Bernoulli's theorem:
$$\frac12\rho v^2 = \rho gh + \Delta P \Rightarrow v^2 = 2\left(10\times0.4 + \frac{600}{1000}\right) = 9.2$$
$v \approx 3.03\ \text{m s}^{-1}$, so $v = 3$.
Solution by Sreeraj P, M.Sc Physics