Question

A hemispherical tank, full of water, is emptied by a pipe at the rate of $\frac{25}{7}$ litres per second. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m? [CBSE 2012]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hemispherical}\mathrm{tank},r=\frac{3}{2}\mathrm{m} \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{hemispherical}\mathrm{tank}=\frac{2}{3}\mathrm{\pi }{r}^3 \\ =\frac{2}{3}\times \frac{22}{7}\times \frac{3}{2}\times \frac{3}{2}\times \frac{3}{2} \\ =\frac{99}{14}{\mathrm{m}}^3 \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{half}\mathrm{tank}=\frac{1}{2}\times \frac{99}{14} \\ =\frac{99}{28}{\mathrm{m}}^3 \\ =\frac{99}{28}\mathrm{kL} \\ =\frac{99000}{28}\mathrm{L} \\ \mathrm{As},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{water}\mathrm{emptied}\mathrm{by}\mathrm{the}\mathrm{pipe}=\frac{25}{7}\mathrm{L}/\mathrm{s} \\ \mathrm{So},\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{empty}\mathrm{half}\mathrm{the}\mathrm{tank}=\frac{\left({\displaystyle \frac{99000}{28}}\right)}{\left({\displaystyle \frac{25}{7}}\right)} \\ =\frac{99000}{25\times 4} \\ =990\mathrm{s} \\ =\frac{990}{60}\min \\ =16.5\min \\ =16\min 30\mathrm{s} \end{gathered}$$

So, the time taken to empty half the tank is 16 minutes and 30 seconds.