A hemispherical tank full of water is emptied by a pipe at the rate of 7.7 liters per second.  How much time (in hours) will it take to empty \(\frac{1}{3}\)rd part of the tank, if the internal radius of the tank is 10.5 m?

\(\frac{175}{6}\) hours

Volume of Hemispherical tank = volume of cylindrical pipe × time

\(\frac{1}{3}\) × \(\frac{2}{3}\)\(\pi \)r³ = \(\frac{1}{3}\) × \(\frac{2}{3}\) ×\(\frac{22}{7}\) × 10.5 × 10.5 × 10.5

= 808.5 m³ 

convert into liters = 808.5 × 1000 liters = 808500

pipe flow = 7.7 × 60 × 60 liter/hour

Time in hours = \(\frac{808500}{7.7 × 60 × 60}\) = \(\frac{175}{6}\) hours