In one fortnight of a given month, there was a rainfall of $ 10 \mathrm{~cm} $ in a river valley. If the area of the valley is $ 7280 \mathrm{~km}^{2} $, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each $ 1072 \mathrm{~km} $ long, $ 75 \mathrm{~m} $ wide and $ 3 \mathrm{~m} $ deep.

Given:

In one fortnight of a given month, there was a rainfall of \( 10 \mathrm{~cm} \) in a river valley.

The area of the valley is \( 7280 \mathrm{~km}^{2} \).

To do:

We have to show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each \( 1072 \mathrm{~km} \) long, \( 75 \mathrm{~m} \) wide and \( 3 \mathrm{~m} \) deep.

Solution:

Height of the rain $=10\ cm$

$=\frac{10}{100}\ m$

$=\frac{10}{100\times1000}\ km$

Total rainfall in the river valley $=$ area of the valley $\times$ height of the rain

$=7280\times\frac{10}{100\times1000}$

$=0.7280\ km^3$

Volume of normal water in each river $=1072\times\frac{75}{1000}\times\frac{3}{1000}$

$= 0.2412\ km^3$

Volume of normal water in 3 rivers $=3\times0.2412\ km^3$

$=0.7236\ km^3$

Therefore, the total rainfall was approximately equivalent to the addition to the normal water of three rivers.

Hence proved.

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Updated on: 10-Oct-2022

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