Question

Question 10
500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is $0.04m^3$ ?

Answer 1

Let the rise of water level in the pond be $h$ m, when $500$ persons are taking a dip into a cuboidal pond.

Given that,
Length of the cuboidal pond $=80$ m
Breadth of the cuboidal pond$=50$ m

Now, volume for the rise of water level in the pond
$=Length\times Breadth\times Height$
$=80\times 50\times h=4000h{m}^3$

And the average displacement of the water by one person
$=0.04m^3$

So, the average displacement of the water by $500$ persons = $500\times 0.04m^3$

Now, by given condition
Volume for the rise of water level in the pond = Average displacement of the water by 500 persons
$\Rightarrow 4000h=500\times 0.04$
$\therefore h=\frac{500\times 0.04}{4000}$
$=\frac{20}{4000}$
$=\frac{1}{200}m$
$=0.005m(1m=100cm)$
$=0.005\times 100cm$
$=0.5cm$

Hence , the required rise of water level in the pond is $0.5cm$.