$A$ hemispherical tank full of water is emptied by a pipe at the rate of $3 \frac{4}{7}$ $litres$ per $second.$ How much time will it take to empty half the tank,if it is $3 \,m$ in $diameter?$ (in $minutes$) (Take $\pi=\frac{22}{7}$)

(16.5) The radius of the hemispherical tank is $r = \frac{3}{2} \, m = 1.5 \, m$.
The total volume of the hemispherical tank is $V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (1.5)^3 = \frac{2}{3} \times \frac{22}{7} \times \frac{27}{8} = \frac{99}{14} \, m^3$.
Since $1 \, m^3 = 1000 \, litres$,the total volume in litres is $\frac{99}{14} \times 1000 = \frac{99000}{14} \, litres$.
We need to empty half of the tank,so the volume to be emptied is $\frac{1}{2} \times \frac{99000}{14} = \frac{99000}{28} \, litres$.
The rate of emptying is $3 \frac{4}{7} = \frac{25}{7} \, litres/second$.
Time taken in seconds $= \frac{\text{Volume}}{\text{Rate}} = \frac{99000}{28} \div \frac{25}{7} = \frac{99000}{28} \times \frac{7}{25} = \frac{99000}{100} = 990 \, seconds$.
To convert into minutes,divide by $60$: $990 \div 60 = 16.5 \, minutes$.