1. Given data

• Radius of cylindrical tank: $r = 1\,\text{m}$
• Water level from bottom: $15\,\text{m}$
• Hole is at height $5\,\text{m}$ from bottom
• Therefore, depth of hole below water surface: $h = 15 - 5 = 10\,\text{m}$
• Applied force on piston: $F = 5\times 10^5\,\text{N}$
• Atmospheric pressure: $P_A = 1.01\times 10^5\,\text{Pa}$
• Density of water: $\rho = 1000\,\text{kg/m}^3$
• $g = 10\,\text{m/s}^2$

2. Pressure applied by piston

Area of top surface: $A = \pi r^2 = \pi(1)^2 = \pi\,\text{m}^2$

Extra pressure due to piston: $P_{\text{piston}} = \frac{F}{A} = \frac{5\times 10^5}{\pi}\,\text{Pa}$

Numerically, $P_{\text{piston}} \approx \frac{5\times 10^5}{3.14} \approx 1.59\times 10^5\,\text{Pa}$

3. Pressure just inside the hole

The pressure at the top surface is atmospheric plus piston pressure: $P_{\text{top}} = P_A + \frac{F}{A}$

At the hole, hydrostatic pressure adds $\rho g h$: $P_{\text{hole,in}} = P_A + \frac{F}{A} + \rho g h$

Outside the hole, pressure is atmospheric: $P_{\text{hole,out}} = P_A$

Hence effective pressure difference driving outflow is: $\Delta P = \frac{F}{A} + \rho g h$

4. Using Bernoulli / efflux formula

For efflux speed, $\frac{1}{2}\rho v^2 = \Delta P$

So, $\frac{1}{2}\rho v^2 = \frac{F}{A} + \rho g h$

$v = \sqrt{\frac{2}{\rho}\left(\frac{F}{A} + \rho g h\right)}$

Substitute values: $v = \sqrt{\frac{2}{1000}\left(\frac{5\times 10^5}{\pi} + 1000\times 10\times 10\right)}$

$v = \sqrt{\frac{2}{1000}\left(1.59\times 10^5 + 1.0\times 10^5\right)}$

$v = \sqrt{\frac{2}{1000}(2.59\times 10^5)}$

$v = \sqrt{518}$

$v \approx 22.8\,\text{m/s}$

5. Check with options

The physically correct calculation gives: $\boxed{v \approx 22.8\,\text{m/s}}$

This does not match any option.

6. Likely source of textbook/key answer

If one incorrectly uses only the piston pressure contribution and ignores the $10\,\text{m}$ hydrostatic head, then: $v = \sqrt{\frac{2(F/A)}{\rho}} = \sqrt{\frac{2(1.59\times 10^5)}{1000}}$ $v = \sqrt{318} \approx 17.8\,\text{m/s}$ which matches option C.

But this is incomplete, because the hole is $10\,\text{m}$ below the water surface, so hydrostatic pressure must also be included.

7. Conclusion

The correct speed should be: $\boxed{22.8\,\text{m/s}}$

Since this is not among the options, the stored answer $\text{C}$ appears to come from neglecting the hydrostatic head.