The area of cross-section of a large tank is $$0.5$$ m$$^2$$. It has a narrow opening near the bottom having area of cross-section $$1$$ 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 ______ cm s$$^{-1}$$.
[Take $$g = 10$$ m s$$^{-2}$$]

A large tank has a cross-section area of $$A = 0.5$$ m$$^2$$, a narrow opening near the bottom with area $$a = 1$$ cm$$^2 = 1 \times 10^{-4}$$ m$$^2$$, a load of 25 kg on the water at the top, and a water height of $$h = 40$$ cm = 0.4 m above the bottom.

At the surface of the water, in addition to atmospheric pressure, there is extra pressure due to the applied load: $$P_{load} = \frac{mg}{A} = \frac{25 \times 10}{0.5} = 500 \text{ Pa}$$

Let the top surface be point 1 and the opening be point 2. Both are exposed to atmospheric pressure $$P_0$$ (at the opening) and $$P_0 + P_{load}$$ (at the surface due to the load). Applying Bernoulli’s equation between these points and neglecting the speed of water at the top surface (since $$A \gg a$$) gives

$$P_0 + P_{load} + \rho g h = P_0 + \frac{1}{2}\rho v^2$$

Rearranging this equation yields

$$P_{load} + \rho g h = \frac{1}{2}\rho v^2$$

Substituting the values, the pressure due to the water column is

$$\rho g h = 1000 \times 10 \times 0.4 = 4000 \text{ Pa}$$

Therefore,

$$500 + 4000 = \frac{1}{2} \times 1000 \times v^2$$

which simplifies to

$$4500 = 500 \times v^2$$

and hence

$$v^2 = 9$$

so that

$$v = 3 \text{ m/s} = 300 \text{ cm/s}$$

The velocity of water coming out of the opening is 300 cm s$$^{-1}$$.