A cylindrical tank of radius $5\text{m}$ is filled with water up to its brim. The water is pumped out through a pipe at the top of the tank. Assuming water density $\rho =1000{\text{kg/m}}^3$ and $g=9.8{\text{m/s}}^2$, find the total work required to empty the tank. (Give your answer in kilojoules, $1\text{kJ}=10^3\text{J}$.)

Step-by-step solution

1. Consider a thin horizontal slice of water at depth $y$ metres below the top surface; its thickness is $dy$.
2. The radius of the tank is constant $r=5\text{m}$, so the area of the slice is $\pi r^2=25\pi {\text{m}}^2$.
3. Volume of the slice: $dV=25\pi dy$ cubic metres.
4. Mass of the slice: $dm=\rho dV=1000\times 25\pi dy=25000\pi dy\text{kg}$.
5. The slice must be lifted a distance $y$ to the top; work on the slice: $dW=dmgy=25000\pi gydy$.
6. Integrate $dW$ from $y=0$ (top) to $y=5$ (bottom): $W=25000\pi g{\int }_0^{5}ydy$.
7. Compute the integral: ${\int }_0^{5}ydy={\left[\frac{y^2}{2}\right]}_0^5=\frac{25}{2}=12.5$.
8. Substitute values: $W=25000\pi \times 9.8\times 12.5$.
9. Calculate: $25000\times 12.5=312500$, then $312500\times 9.8=3,062,500$, and $3,062,500\pi \approx 9,620,000\text{J}$.
10. Convert joules to kilojoules: $9.62\times 10^6\text{J}=9620\text{kJ}$ (rounded to three significant figures).
11. Hence the work required is $\boxed{9.62\times 10^6\text{J}=9620\text{kJ}}$.

Concept tested: Work done in pumping fluids

Published 8 September 2026 · How this solution was prepared and checked