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 0.04 m3?
Let the rise of water level in the pond be hm, when 500 persons are taking a dip into a cuboidal pond.
Given that
Length of the cuboidal pond $=80 \mathrm{~m}$
Breadth of the cuboidal pond $=50 \mathrm{~m}$
Now, volume for the rise of water level in the pond
$=$ Length $\times$ Breadth $\times$ Height
$=80 \times 50 \times h=4000 h \mathrm{~m}^{3}$
and the average displacement of the water by a person $=0.04$
$\mathrm{~m}^{3}$
So, the average displacement of the water by 500 persons $=500 \times 0.04 \mathrm{~m}^{3}$
Now, by given condition,
Volume for the rise of water level in the pond = Average displacement of the water by
500 persons
$\Rightarrow$ $4000 h=500 \times 0.04$
$\therefore$ $h=\frac{500 \times 0.04}{4000}=\frac{20}{4000}=\frac{1}{200} \mathrm{~m}$
$=0.005 \mathrm{~m}$
$=0.005 \times 100 \mathrm{~cm}$
$=0.5 \mathrm{~cm}$
Hence, the required rise of water level in the pond is 0.5 cm.
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