Water flows out through a circular pipe whose internal diameter is $2\ cm$, at the rate of $6$ metres per second into a cylindrical tank. The radius of whose base is $60\ cm$. Find the rise in the level of water in $30$ minutes?
Given:

Water flows out through a circular pipe whose internal diameter is $2\ cm$, at the rate of $6$ metres per second into a cylindrical tank. The radius of whose base is $60\ cm$.

To do:

We have to find the rise in the level of water in $30$ minutes

Solution:

Internal diameter of the pipe $= 2\ cm$

This implies,

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

$= 1\ cm$

Speed of the water flow $= 6\ m$ per second

Water in 30 minutes $(h) = 6 \times 60 \times 30\ m$

$= 10800\ m$

Therefore,

Volume of water $= \pi r^2h$

$=\frac{22}{7} \times \frac{1}{100} \times \frac{1}{100} \times 10800$

$=\frac{22 \times 10800}{100 \times 100 \times 7}$

Radius of the base of the cylindrical tank $(\mathrm{R})=60 \mathrm{~cm}$

Let the height of the water be $\mathrm{H}$.

This implies,

$\pi \mathrm{R}^{2} \mathrm{H}=\frac{22 \times 10800}{7}$

$\frac{22}{7} \times \frac{60}{100} \times \frac{60}{100} \mathrm{H}=\frac{22 \times 10800}{7}$

$\mathrm{H}=\frac{22 \times 10800 \times 7 \times 100 \times 100}{7 \times 22 \times 60 \times 60 \times 100 \times 100}$

$H=\frac{30000}{10000}$

$H=3 \mathrm{~m}$

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