Question

$500$ persons are taking a dip into a cuboidal pond which is $80\mathrm{m}$ long and $50\mathrm{m}$ broad. What is the rise of water level in the pond, if the average displacement of the water by a person is $0.04{\mathrm{m}}^3$?

Answer 1

Given

Average displacement by a person $=0.04{\mathrm{m}}^3$

Dimensions of pond

Length of the pond, $\mathrm{l}=80\mathrm{m}$

The breadth of the pond, $\mathrm{b}=50\mathrm{m}$

Number of persons taking a dip $=500$

To Find out

We have to determine the rise of the water level in the pond.

Solution:

Average displacement by $500$ persons $=500\times 0.04=20{\mathrm{m}}^3$

Hence, the volume of water raised in the pond $=20{\mathrm{m}}^3$

The volume of water raised in a pond $=80\left(50\right)\left(\mathrm{h}\right)$

$\Rightarrow 20{\mathrm{m}}^3=4000\times \mathrm{h}$

$\Rightarrow \mathrm{h}=0.005\mathrm{m}=0.5\mathrm{cm}$

Therefore, a rise in the height of water $=0.5\mathrm{cm}.$

Hence, the rise of water level in the pond$=0.5\mathrm{cm}$