Question

a farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field , which is 10m in diameter and 2 m deep. If water flows through the pipe at the rate of 3km/hr , in how much time the tank be filled ?

Answer 1

Solution:-

Speed of water = 3 km/hr
= 3000/60
= 50 meter
And, 
Internal radius of pipe = 20/2 = 10 cm or 0.1 m
So,
Area of the cross section = πr² 
= π(1/10)²
= 0.01π m²
Volume of the water that flows in 1 minute
= 0.01π × 50
= 0.5π m³
Let the time taken to fill the tank = t minutes 
So, Volume of the water that flows in t minutes = 0.5πt m³
And,
Radius of the base of the tank = 10/2 = 5 m
Height = 2 m
Volume of the cylindrical tank = πr²h
= π*5*5*2
= 50π m³
So,
Volume of the water that flows in t minutes = Volume of the tank
⇒ 0.5πt = 50π 
⇒ t = 50/0.5
t = 100 minutes 
So, the tank will be filled in 100 minutes.
Answer.

Answer 2

Question :-

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his feild, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled.

$\huge\boxed{ \underline{ \bf{ \bigstar \: Solution}}}$

Given, speed of flow of water = 3 km/h = 3 × 1000 m / h

$\longrightarrow \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [ \because \bf \: 1km \: = \: 1000 \: m]$

$\therefore \: \: \: \rm \: Length \: of \: water \: flow \: in \: 1 \: h \: = 3000 \: m$

$\rm \: Now, \: area \: of \: the \: pipe \: which \: is \: in \: the \\ \rm form \: of \: a \: circle$

$= \: (10) {}^{2} \pi \bigg[ \because \rm \: area \: of \: circle \: = \pi \: r {}^{2} \: and \: r = \frac{20}{2} \: cm = 10 \: cm \bigg]$

$\rm \: = \: 100\pi \: cm {}^{2} = \frac{\pi}{100} \: m {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg[ \because \: 1 \: cm = \frac{1}{100} \: m \bigg]$

$\rm \: For \: cylindrical \: tank,$

$\rm \: Diameter \: = 10 \: cm$

$\therefore \: \: \: \rm \: Radius \: = \frac{10}{2} \: m \: and \: height \: = 2 \: m, \: then$

$\rm \: Volume \: of \: cylindrical \: tank \: = \pi \times \bigg( \frac{10}{2} \bigg){}^{2} \times 2 \: [ \because \: v = \pi \: r {}^{2} h]$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm = \: 50\pi \: m {}^{3}$

$\therefore \: \: \: \rm \: Required \: time$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm = \: \frac{volume \: of \: cylindrical \: tank}{area \: of \: the \: pipe \times \: length \: of \: water \: flow \: in \: 1 \: h}$

$\: \: \: \: \: \: \: \rm = \: \: \frac{50\pi}{ \frac{\pi}{100} \times 3000} \: h$

$\: \: \: \: \: \: \: \rm = \: \frac{50 \times 60 \times 100}{3000} \: min$

$\: \: \: \: \: \: \: \: \: \rm = \: 100 \: min$

$\rm \: Hence, \: in \: 100 \: min, \: the \: tank \: will \: be \: filled.$