Question

A pool has a uniform circular cross-section of radius $5m$ and uniform depth $1.4m$. It is field by a pipe which develops water at the ratio of $50L/s$. Calculate in minutes, the time taken field the pool. If the pool is emptied in $42$ minutes by another cylindrical pipe through which water flaws at the rate of $2m/s$, calculate the radius of the pipe in centimeter.

Answer 1

Volume of pool $=\pi R^{2}h=\pi \times 5\times 5\times 1.4=35\pi m^3=3500\pi$ litres

Volume of water coming out $=50L$ per second

Time taken to fill the tank $=\frac{35000\pi }{50}sec$

$=\frac{35000\times 22}{50\times 7\times 60}$ min

$=36.67$ min

Let radius of pipe be ${}^{\prime }{r}^{\prime }$ cm. Then

Volume of water coming out from pipe in 1 sec, $=({\pi r}^2\times 200){cm}^3$

Time required to empty $=42min=42\times 60$ sec

$\Rightarrow \frac{35\times {10}^6{cm}^3\times \pi }{{\pi r}^2\times 200}=42\times 60$

$\Rightarrow r^2=\frac{35\times {10}^6}{42\times 60\times 200}=69.44cm$

$\Rightarrow r=\sqrt{69.44}=8.33cm$