A farmer runs a pipe of internal diameter 20 cm from

The internal radius of the pipe is $10 \mathrm{~cm}=0.1 \mathrm{~m}$. The water is flowing in the pipe at $3 \mathrm{~km} / \mathrm{hr}=3000 \mathrm{~m} / \mathrm{hr}$. Let the cylindrical tank will be filled in $t$ hours. Therefore, the length of the flowing water in $t$ hours is $=3000 \times t$ meter

Therefore, the volume of the flowing water is

$V_{1}=\pi \times(0.1)^{2} \times 3000 \times t \mathrm{~m}^{3}$

The radius of the cylindrical tank is 5 m and the height is 2 m. Therefore, the volume of the cylindrical tank is

$V_{2}=\pi \times(5)^{2} \times 2 \mathrm{~m}^{3}$

Since, we have considered that the tank will be filled in t hours; therefore the volume of

the flowing water in t hours is same as the volume of the cylindrical tank. Hence, we have

$V_{1}=V_{2}$

$\Rightarrow \pi \times(5)^{2} \times 2=\pi \times(0.1)^{2} \times 3000 \times t$

$\Rightarrow t=\frac{(5)^{2} \times 2}{(0.1)^{2} \times 3000}$

$\Rightarrow t=\frac{5}{3}$ hours

$t=\frac{5 \times 60}{3}=100$ minut

Hence, the tank will be filled in 1 hour 40 minutes