Water flows at the rate of $ 15 \mathrm{~km} / \mathrm{hr} $ through a pipe of diameter $ 14 \mathrm{~cm} $ into a cuboidal pond which is $ 50 \mathrm{~m} $ long and $ 44 \mathrm{~m} $ wide. In what time will the level of water in the pond rise by $ 21 \mathrm{~cm} $?

Given:

Water flows at the rate of \( 15 \mathrm{~km} / \mathrm{hr} \) through a pipe of diameter \( 14 \mathrm{~cm} \) into a cuboidal pond which is \( 50 \mathrm{~m} \) long and \( 44 \mathrm{~m} \) wide.

To do:

We have to find the time it will take for the level of water in the pond to rise by \( 21 \mathrm{~cm} \).

Solution:

Length of the pond $= 50\ m$

Width of the pond $= 44\ m$
Level of water in the pond $=21 \mathrm{~cm}$

$=\frac{21}{100} \mathrm{~m}$

Volume of water in the pond $=(50 \times 44 \times \frac{21}{100})^{3}$

$=462 \mathrm{~m}^{3}$

Radius of the pipe $=7 \mathrm{~cm}$

$=\frac{7}{100} \mathrm{~m}$

Rate of water flowing through the pipe $=15\ km/hr$

$=15000\ m/hr$

Volume of water flow in $1 \mathrm{~hr}=\pi \mathrm{R}^{2} \mathrm{H}$

$=(\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 15000)$

$=231 \mathrm{~m}^{3}$
This implies,

Time taken for $1 \mathrm{~m}^{3}$ of water to fall in the pond $=\frac{1}{231}$ hr. Therefore,

Time taken for $462 \mathrm{~m}^{3}$ of water to fall in the pond $=(\frac{1}{231} \times 462$

$=2 \mathrm{~hr}$
The required time is 2.