A pool has a uniform circular cross section of radius 5 m and uniform depth 1.4 m. It is filled by a pipe which delivers water at the rate of 20 litres per sec. Calculate, in minutes, the time taken to fill the pool. If the pool is emptied in 42 min. by another cylindrical pipe through which water flows at 2 m per sec, calculate the radius of the pipe in cm.

Given,

Radius of pool (R) = 5 m,

Depth of pool (H) = 1.4 m.

Since, pool is in the form of cylinder.

Volume of water in pool = πR²H

= $\dfrac{22}{7} \times 5^2 \times 1.4$ = 110 m³

= 110 × 1000 litres = 11 × 10⁴ liters.

Volume of water coming out of filling pipe = 20 litres/sec.

Time taken to fill the tank = $\dfrac{11 \times 10^4}{20}$ second

= $\dfrac{11 \times 10^4}{20 \times 60}$ minutes

= $\dfrac{1100}{12} = 91\dfrac{2}{3}$ minutes.

Let radius of pipe be r cm.

Volume of water coming out from draining pipe in 1 sec = Area of cross section × Rate

= πr² × 2 m/sec

= (200 × πr²) cm³

= (200 × πr² × 10^-3) liters.

Given,

Time required for emptying the pool = 42 min = 42 × 60 seconds.

$\therefore \dfrac{\text{Volume of water in pool}}{\text{Volume of water drained out in 1 sec}} = 42 \times 60 \\[1em] \Rightarrow \dfrac{11 \times 10^4}{200 \times 10^{-3} \times πr^2} = 42 \times 60 \\[1em] \Rightarrow \dfrac{11 \times 10^2 \times 10^3}{2 \times \dfrac{22}{7} \times r^2} = 42 \times 60 \\[1em] \Rightarrow \dfrac{1100 \times 7 \times 10^3}{44r^2} = 2520 \\[1em] \Rightarrow \dfrac{100 \times 7 \times 10^3}{4r^2} = 2520 \\[1em] \Rightarrow \dfrac{700000}{4r^2} = 2520 \\[1em] \Rightarrow r^2 = \dfrac{700000}{4 \times 2520} \\[1em] \Rightarrow r^2 = \dfrac{100000}{4 \times 360} \\[1em] \Rightarrow r^2 = \dfrac{2500}{36} \\[1em] \Rightarrow r = \sqrt{\dfrac{2500}{36}} \\[1em] \Rightarrow r = \dfrac{50}{6} = \dfrac{25}{3} = 8\dfrac{1}{3} \text{ cm}.$

Hence, time taken to fill the pool = $91\dfrac{2}{3}$ minutes and radius of pipe = $8\dfrac{1}{3}$ cm.