In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, 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.
From the question, it is clear that
Total volume of 3 rivers = 3 × [(Surface area of a river) × Depth]
Surface area of a river = [1072×(75/1000)] km
Depth = (3/1000) km
Volume of 3 rivers = 3 × [1072 × (75/1000)] × (3/1000)
= 0.72 km3
Now, volume of rainfall = total surface area × total height of rain
= \(9780 \times \frac{10}{100 \times 1000}\)
= 9.7 km3
For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.
= 9.7 km3 ≠ 0.72 km3
So, the question statement is false.