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 m3
= 45000 m3
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 m2 = 56.25 hectares.