In one fortnight of a given month, there was a rainfall of 10cm in a river valley. If the area of the valley is 97280 sq.km , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
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Solution:-
Area of the valley = 97280 sq km
Rise in the level of water in the valley = h = 10 cm (10/100000) km = (1/10000)km
Thus, amount of rainfall = A*h = 97280*(1/10000)
= 9.728 km³
In 14 days, rainfall = 9.728 km³
In 1 day, rainfall = 9.728/14
= 0.6948 km³
Volume of water in each river = L*B*H
= 1072*75/1000*3/1000
= 0.2412 km³
So volume of three rivers = 3*0.2412
= 0.7236 km³
So, amount of rainfall = 0.6948 km³ is not exactly equal but approximately equal to the water in three rivers.
Area of the valley = 97280 sq km
Rise in the level of water in the valley = h = 10 cm (10/100000) km = (1/10000)km
Thus, amount of rainfall = A*h = 97280*(1/10000)
= 9.728 km³
In 14 days, rainfall = 9.728 km³
In 1 day, rainfall = 9.728/14
= 0.6948 km³
Volume of water in each river = L*B*H
= 1072*75/1000*3/1000
= 0.2412 km³
So volume of three rivers = 3*0.2412
= 0.7236 km³
So, amount of rainfall = 0.6948 km³ is not exactly equal but approximately equal to the water in three rivers.
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