Question

Question 4
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is $97280k{m}^2$, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer 1

Area of the valley $=97280k{m}^2$
If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley $\times 10$ cm
Amount of rainfall in the valley$=97280k{m}^2\times 10cm$
$=97280\times (1000m{)}^2\times \frac{10}{100}m$
$=9728\times {10}^6{m}^3$
$=9.728\times {10}^9{m}^3$
Length of each river, $l=1072km=1072\times 1000m=1072000m$
Breadth of each river, $b=75m$
Depth of each river, $h=3m$
Volume of each river $=l\times b\times h$
$=1072000\times 75\times 3m^3$
$=2.412\times {10}^8{m}^3$
Volume of three such rivers = 3 $\times$ Volume of each river
$=3\times 2.412\times {10}^8{m}^3$
$=7.236\times {10}^8{m}^3$
Thus, the total rainfall is not approximately same as the volume of the three rivers.