A hemispherical tank, full of water, is emptied by a pipe at the rate of

We have,

the radius of the hemispherical $\operatorname{tank}, r=\frac{3}{2} \mathrm{~m}$

Volume of the hemispherical $\operatorname{tank}=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}$

$=\frac{99}{14} \mathrm{~m}^{3}$

Now,

Volume of half $\operatorname{tank}=\frac{1}{2} \times \frac{99}{14}$

$=\frac{99}{28} \mathrm{~m}^{3}$

$=\frac{99}{28} \mathrm{~kL}$

$=\frac{99000}{28} \mathrm{~L}$

As, the rate of water emptied by the pipe $=\frac{25}{7} \mathrm{~L} / \mathrm{s}$

So, the time taken to empty half the tank $=\frac{\left(\frac{99000}{28}\right)}{\left(\frac{25}{7}\right)}$

$=\frac{99000}{25 \times 4}$

$=990 \mathrm{~s}$

$=\frac{990}{60} \mathrm{~min}$

$=16.5 \mathrm{~min}$

$=16 \mathrm{~min} 30 \mathrm{~s}$

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