1. Given data

• Area of tank cross-section: $A = 0.5\,\text{m}^2$
• Area of opening: $a = 1\,\text{cm}^2 = 10^{-4}\,\text{m}^2$
• Load applied on top of water: $m = 25\,\text{kg}$
• Height of water above opening: $h = 40\,\text{cm} = 0.4\,\text{m}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$
• Density of water: $\rho = 1000\,\text{kg m}^{-3}$

Since the tank is large, speed of water at the top surface is neglected.

2. Pressure applied by the load

The extra pressure on the water surface due to the load is $P = \frac{mg}{A} = \frac{25\times 10}{0.5} = 500\,\text{Pa}$

This corresponds to an equivalent water column height $h_p = \frac{P}{\rho g} = \frac{500}{1000\times 10} = 0.05\,\text{m}$

3. Total effective head

The water already has hydrostatic head due to height $h = 0.4\,$m, and the load adds another $0.05\,$m.

So total head is $H = 0.4 + 0.05 = 0.45\,\text{m}$

4. Apply Bernoulli/Torricelli theorem

Velocity of efflux is $v = \sqrt{2gH}$

Substitute values: $v = \sqrt{2\times 10\times 0.45} = \sqrt{9} = 3\,\text{m s}^{-1}$

5. Convert to cm/s

$3\,\text{m s}^{-1} = 300\,\text{cm s}^{-1}$

6. Final answer

$\boxed{300}$