1. Given data

• Cross-sectional area of tank:
  $A = 0.4\,\text{m}^2$
• Area of opening near bottom:
  $a = 1\,\text{cm}^2 = 10^{-4}\,\text{m}^2$
• Load applied on top:
  $m = 24\,\text{kg}$
• Height of water above opening:
  $h = 40\,\text{cm} = 0.4\,\text{m}$
• Acceleration due to gravity:
  $g = 10\,\text{m s}^{-2}$

2. Pressure applied by the load

The load on the top surface produces extra pressure on the water:

$P_{\text{extra}} = \frac{mg}{A}$

Substituting:

$P_{\text{extra}} = \frac{24\times 10}{0.4} = \frac{240}{0.4} = 600\,\text{Pa}$

3. Equivalent water head due to this pressure

Using

$P = \rho g h_{\text{eq}}$

so

$h_{\text{eq}} = \frac{P_{\text{extra}}}{\rho g} = \frac{600}{1000\times 10} = 0.06\,\text{m}$

4. Total effective head

The outlet speed is due to:

• actual water height: $0.4\,\text{m}$
• extra equivalent head from the load: $0.06\,\text{m}$

Hence,

$H = 0.4 + 0.06 = 0.46\,\text{m}$

5. Apply Bernoulli/Torricelli theorem

Since tank area is much larger than outlet area,

$v = \sqrt{2gH}$

Thus,

$v = \sqrt{2\times 10\times 0.46} = \sqrt{9.2}$

$v \approx 3.03\,\text{m s}^{-1}$

6. Nearest integer

$v \approx 3$

Therefore, the required integer answer is:

$\boxed{3}$