1. Given data

• Height of tank: $H = 1.6\,\text{m}$
• Cross-sectional area of tank: $A = 0.5\,\text{m}^2$
• Hole is at height $0.9\,\text{m}$ from bottom
• Mass placed on top surface: $m = 50\,\text{kg}$
• $g = 10\,\text{m/s}^2$
• Density of water: $\rho = 1000\,\text{kg/m}^3$

2. Depth of hole below the top surface

Since the tank is completely full, the top water surface is at height $1.6\,\text{m}$.

So the hole is below the top surface by

$h = 1.6 - 0.9 = 0.7\,\text{m}$

3. Extra pressure due to the applied load

The load on the top surface produces additional pressure

$P_{\text{extra}} = \frac{mg}{A} = \frac{50 \times 10}{0.5} = 1000\,\text{Pa}$

4. Pressure just inside the hole

Pressure at the hole inside the tank is due to:

• atmospheric pressure,
• hydrostatic pressure of water column of depth $0.7\,\text{m}$,
• extra pressure due to the load.

Thus,

$P_{\text{in}} = P_{\text{atm}} + \rho g h + P_{\text{extra}}$

Outside the hole,

$P_{\text{out}} = P_{\text{atm}}$

So the effective pressure difference causing outflow is

$\Delta P = \rho g h + P_{\text{extra}}$

$\Delta P = 1000 \times 10 \times 0.7 + 1000 = 7000 + 1000 = 8000\,\text{Pa}$

5. Use Bernoulli/Torricelli relation

For efflux speed,

$\frac{1}{2}\rho v^2 = \Delta P$

So,

$\frac{1}{2}(1000)v^2 = 8000$

$500v^2 = 8000$

$v^2 = 16$

$v = 4\,\text{m/s}$

6. Check options

• A: $2\,\text{m/s}$
• B: $5\,\text{m/s}$
• C: $3\,\text{m/s}$
• D: $4\,\text{m/s}$

Hence the correct option is D.