Ex.13.3 Q8 Surface Areas and Volumes Solution - NCERT Maths Class 10

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Question

Water in canal, \(6\,\rm{m}\) wide and \(1.5\,\rm{m}\) deep, is flowing with a speed of \(10\,\rm{km/h}\). how much area will it irrigate in \(30\) minutes, if \(8\,\rm{cm}\) of standing water is needed?

Text Solution

 

What is known?

Width of the canal is \(6\rm{m}\) and depth is \(1.5\rm{m}\) and speed of flowing water is \(10\rm{km/h}\)

Height of the standing water is \(8\rm{cm}\) in the area to be irrigated for \(30\) minutes.

What is unknown?

The area to be irrigated in \(30\) minutes with \(8\rm{cm}\) of standing water.

Reasoning:

Draw a figure to visualize the shapes better

From the figure it’s clear that shape of the cross-section of the canal is cuboidal. So, volume of the water, flowing at the speed of \(10\rm{km/h}\) for \(30\) minutes will be same as the volume of water to irrigate the area with \(8\rm{cm}\) of standing water.

To find the volume of the water we need to find length of the water flowing through canal in \(30\) minutes at the speed of \(10\rm{km/h}\)

Hence, length of the water flowing through canal in \(1\) hour \(= 10\rm{km}\)

Length of the water flowing through canal in \(30\) minutes \(= 5\rm{km} = 5000\rm{m}\)

 We will find the volume of the water by using formula;

Volume of the cuboid \( = lbh\)

where \(l, b\) and \(h\) are length, breadth and height of the cuboid respectively.

Volume of the water to irrigate the area \(=\) Area to be irrigated \(\times\) height of the standing water

Volume of water flowing through the canal in \(30\) minutes \(=\) Area to be irrigated \(\times\) height of the standing water

Width of the cuboidal canal, \(b = 6m\)

Depth of the cuboidal canal,\(h = 1.5m\)

Speed of water flowing through the canal is \(10\rm{km/h}\)

Length of the water flowing through canal in \(1\) hour \((60\rm{minutes)}= 10\rm{km}\) Length of the water flowing through canal in \(30\) minutes\(\begin{align} = \frac{{30}}{{60}} \times 10km = 5km\end{align}\)

\[l = 5 \times 1000m = 5000m\]

Height of the standing water,\(\begin{align}{h_1} = 8cm = \frac{8}{{100}}m = 0.08m\end{align}\)

Volume of water flowing through the canal in\( 30\,\rm{minutes} =\) Area to be irrigated  \(\times\) height of the standing water

\(lbh = \) Area to be irrigated \(\times {\rm{ }}{h_1}\)

Area to be irrigated \( \begin{align}= \frac{{lbh}}{{{h_1}}}\end{align}\)

\[\begin{align}&= \frac{{5000m \times 6m \times 1.5m}}{{0.08m}}\\&= 562500{m^2}\end{align}\]

Therefore, area irrigated in \(30 \) minutes is \(562500\text{ m}^2.\)