In chapter surface areas and volumes of ncert class 10 , please help me to solve the exercise 13.3 question number 8 . Find the best solution of this question also give me the easiest and simplest way to solve this tricky question .Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

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# Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? Q.8

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It is given that the canal is the shape of a cuboid with dimensions as:

Breadth (b) = 6 m and Height (h) = 1.5 m

It is also given that

The speed of canal = 10 km/hr

Length of canal covered in 1 hour = 10 km

Length of canal covered in 60 minutes = 10 km

Length of canal covered in 1 min = (1/60)x10 km

Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m

We know that the canal is cuboidal in shape. So,

Volume of canal = lxbxh

= 5000x6x1.5 m

^{3}= 45000 m

^{3}Now,

Volume of water in canal = Volume of area irrigated

= Area irrigated x Height

So, Area irrigated = 56.25 hectares

∴ Volume of canal = lxbxh

45000 = Area irrigatedx8 cm

45000 = Area irrigated x (8/100)m

Or, Area irrigated = 562500 m

^{2 }= 56.25 hectares.