Question : A hemispherical tank full of water is emptied by a pipe at the rate of 7.7 litres per second. How much time (in hours) will it take to empty $\frac{2}{3}$ part of the tank, if the internal radius of the tank is 10.5 m?
Option 1: $\frac{185}{3}$
Option 2: $\frac{185}{6}$
Option 3: $\frac{175}{3}$
Option 4: $\frac{175}{2}$
Correct Answer: $\frac{175}{3}$
Solution :
The volume of the hemispherical tank = $\frac{2}{3}\pi r^3$
= $\frac{2}{3}\times\frac{22}{7}×10.5 × 10.5 × 10.5$
= 2425.5 m
3
The capacity of the tank is = 2425.5 × 1000 L
= 2425500 L
Time taken by the pipe emptied $\frac{2}{3}$ part of the tank is
= $\frac{\frac{2}{3} × 2425500}{7.7}$ sec
= 210,000 sec
Time in hours = $\frac{210,000}{3600}$
= $\frac{175}{3}$ hours
Hence, the correct answer is $\frac{175}{3}$.
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