Ex.13.5 Q4 Surface Areas and Volumes Solution - NCERT Maths Class 10

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Question

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

   

Text Solution

 

What is known?

In one fortnight there was a rainfall of \(10\,\rm{ cm}\) in a river valley of area \( 7280 \,\rm{km^2}\)

What is unknown?

The total rainfall was approximately equivalent to the addition to the normal water of three rivers each \(1072\,\rm{ km} \) long, \(75 \rm{m}\) wide and \(3 \rm{m}\) deep.

Reasoning:

Since there was a rainfall of \(10\,\rm{cm}\) in a river valley of area \(7280\,\rm{ km^2}\) the volume of the rainfall will be calculated by

Volume of the rainfall \(=\) Area of the river valley \(\times\) height of rainfall in the river valley

Since the dimensions of three rivers are known, we can calculate the volume by

Volume of the river \(\begin{align} = \rm{length \times breadth \times height}\end{align}\)

Steps:

Area of the Valley

\[\begin{align}A &= 7280k{m^2} \\&= 7280 \times 1000000{m^2} \\&= 7.28 \times {10^9}{m^2}\end{align}\]

Height of rainfall in a fortnight

\[\begin{align}h& = 10cm\\&= \frac{{10}}{{100}}m \\&= 0.1m\end{align}\]

Volume of the rainfall \(=\) Area of the river valley \(\times\) height of rainfall in the river valley

\[\begin{align}&= 7.28 \times {10^9}{m^2} \times 0.1m\\& = 7.28 \times {10^8}{m^3}\end{align}\]

Length of river, \(\begin{align}l = 1072km = 1072 \times 1000m = 1.072 \times {10^6}m\end{align}\)

Width of river,\(b = 75m\)

Depth of river,\(h = 3m\)

Volume of \(3\) rivers \(= 3\) \(\times\) volume of \(1\) river

\[\begin{align}&= 3lbh\\&= 3 \times 1.072 \times 106m \times 75m \times 3m\\&= 723.6 \times {10^6}{m^3}\\&= 7.236 \times {10^8}{m^3}\end{align}\]

Since\(7.236 \times {10^8}{m^3}\) is approximately equivalent to \(7.28 \times {10^8}{m^3}\)

Therefore, we can say that total rainfall in the valley was approximately equivalent to the addition of normal water of three rivers each \(1072 \,\rm{km}\) long, \(75 \,\rm{m}\) wide and \(3\,\rm{m}\) deep.

 

  
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