Question

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.

Answer 1

Step 1: Simplify the given data

It is given that,

The depth of rainfall $=10\mathrm{cm}=\frac{10}{100000}\mathrm{km}=0.0001\mathrm{km}$ $\left[\because 1\mathrm{km}=1000,00\mathrm{cm}\right]$

The area of the valley $=7280{\mathrm{km}}^2$

The length of each river $=1072\mathrm{km}$

The width of each river $=75\mathrm{m}=\frac{75}{1000}\mathrm{km}=0.075\mathrm{km}$ $[\because 1\mathrm{km}=1000m]$

And, the depth of each river $=3\mathrm{m}=\frac{3}{1000}\mathrm{km}=0.003\mathrm{km}$ $[\because 1\mathrm{km}=1000m]$

Step 2: Calculate the volume of the rainfall

The volume of the rainfall in the valley can be calculated as,

The volume of the rainfall in the valley $=$ area of the valley $\times$ depth of rainfall

$\Rightarrow$ The volume of the rainfall in the valley $=7280{\mathrm{km}}^2\times 0.0001\mathrm{km}$

$\Rightarrow$ The volume of the rainfall in the valley $=7280\times 0.0001{\mathrm{km}}^3$

$\Rightarrow$ The volume of the rainfall in the valley $=0.7280{\mathrm{km}}^3$ …(i)

Step 3: Calculate the volume of the three rivers

The volume of one river can be calculated as,

The volume of one river $=$ length of river $\times$ width of river $\times$ depth of river

$\Rightarrow$ The volume of one river $=1072\mathrm{km}\times 0.075\mathrm{km}\times 0.003\mathrm{km}$

$\Rightarrow$ The volume of one river $=1072\times 0.075\times 0.003{\mathrm{km}}^3$

$\Rightarrow$ The volume of one river $=0.2412{\mathrm{km}}^3$

Then, the volume of three such rivers $=3\times 0.2412{\mathrm{km}}^3$

$\Rightarrow$ The volume of three such rivers $=0.7236{\mathrm{km}}^3$ …(ii)

Now, from equation (i) and equation (ii), it can be concluded that the total rainfall was approximately equivalent to the sum of the volume of water in three rivers.

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