# 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.

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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.

Updated on 10-Oct-2022 13:47:38