The area of cross-section of a large tank is $0.5\ \text{m}^2$. It has a narrow opening near the bottom having area of cross-section $1\ \text{cm}^2$. A load of $25$ kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $40$ cm above the bottom, will be ______ $\text{cm s}^{-1}$.
[Take $g = 10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 300
The load adds a pressure on the water surface:
$$\Delta P = \frac{25\times10}{0.5} = 500\ \text{Pa}$$
Bernoulli's equation between the top surface (speed neglected) and the opening:
$$\frac12\rho v^2 = \Delta P + \rho g h$$
$$v^2 = \frac{2(500)}{1000} + 2(10)(0.4) = 1 + 8 = 9$$
$$v = 3\ \text{m s}^{-1} = 300\ \text{cm s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics