A cylindrical water tank of diameter $1.4\ m$ and height $2.1\ m$ is being fed by a pipe of diameter $3.5\ cm$ through which water flows at the rate of $2$ metre per second. In how much time the tank will be filled?
Given:

A cylindrical water tank of diameter $1.4\ m$ and height $2.1\ m$ is being fed by a pipe of diameter $3.5\ cm$ through which water flows at the rate of $2$ metre per second.

To do:

We have to find the time in which the tank will be filled.

Solution:

Diameter of the cylindrical tank $= 1.4\ m$

This implies,

Radius $(r) = \frac{1.4}{2}$

$=0.7\ m$

Height of the tank $(h) = 2.1\ m$

Therefore,

Volume of the water in the tank $= \pi r^2h$

$\frac{22}{7} \times 0.7 \times 0.7 \times 2.1 \mathrm{~m}^{3}=3.234 \mathrm{~m}^{3}$

Diameter of the tap $=3.5 \mathrm{~cm}$

Radius of the tap $=\frac{3.5}{2}$

$=1.75 \mathrm{~cm}$

Rate of the flow of water in the pipe $=2 \mathrm{~m} / \mathrm{s}$

Volume of the water flow in 1 second $=\pi r^{2} h$

$=\frac{22}{7} \times 1.75 \times 1.75 \times 200$

$=\frac{22}{7} \times \frac{175}{100} \times \frac{175}{100} \times 200$

$=\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 200$

$=1925 \mathrm{~cm}^3$

Time taken to fill the tank $=\frac{3.234 \times 100 \times 100 \times 100}{1925} \mathrm{sec}$

$=\frac{3234 \times 100 \times 100 \times 100}{1000 \times 1925} \mathrm{sec}$

$=1680 \mathrm{sec}$

$=\frac{1680}{60}$ minutes

$=28$ minutes

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