
In one fortnight of a given month, there was a rainfall of 10cm in a river valley. If the area of the valley is 7280 $k{m^2}$, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75m wide and 3m deep.
Answer
597.6k+ views
Hint: In this question, we will use the concept of conversion of solid from one shape to another. We know that when one solid is converted into another its volume remains the same. Using this concept in the above problem we can easily find our required answer.
Complete step by step solution:
Now from the question, we have
Given:
Area of the valley = 7280$k{m^2}$
Volume of rainfall in the area = 7280 $ \times $ 10 $ \times 0.00001$ = 0.728$k{m^3}$
Volume of water in three rivers = $3 \times l \times b \times h = 3 \times 1072 \times 75 \times 0.001 \times 3 \times 0.001 = 0.723k{m^3}$ (converted m = km)
As the volume of both the rainfall and the volume of the three rivers is nearly equal, therefore the total rainfall was approximately equivalent to the addition to the normal water of three rivers.
Note: The volume of the rainfall in the area is found by using the formula
= area of the valley $ \times $ height of rainfall
= 7280 $ \times $ 10 $ \times $ 0.00001 = 0.728$k{m^3}$
Each and every solid that exists occupies some volume. When you convert one solid shape to another, its volume remains the same, no matter how different the new shape is. In fact, if you melt one big cylindrical candle to 5 small cylindrical candles, the sum of the volumes of the smaller candles is equal to the volume of the bigger candle. Hence, when you convert one solid shape to another, all you need to remember is that the volume of the original, as well as the new solid, remains the same.
There are some solid shapes whose volume is given here:
Volume of cylinder = $\pi {r^2}h$
Volume of cube = ${(side)^3}$
Volume of cuboid = $l \times b \times h$
Volume of sphere = $\dfrac{4}{3}\pi {r^3}$
These formulas should be kept in mind for solving these types of questions.
Complete step by step solution:
Now from the question, we have
Given:
Area of the valley = 7280$k{m^2}$
Volume of rainfall in the area = 7280 $ \times $ 10 $ \times 0.00001$ = 0.728$k{m^3}$
Volume of water in three rivers = $3 \times l \times b \times h = 3 \times 1072 \times 75 \times 0.001 \times 3 \times 0.001 = 0.723k{m^3}$ (converted m = km)
As the volume of both the rainfall and the volume of the three rivers is nearly equal, therefore the total rainfall was approximately equivalent to the addition to the normal water of three rivers.
Note: The volume of the rainfall in the area is found by using the formula
= area of the valley $ \times $ height of rainfall
= 7280 $ \times $ 10 $ \times $ 0.00001 = 0.728$k{m^3}$
Each and every solid that exists occupies some volume. When you convert one solid shape to another, its volume remains the same, no matter how different the new shape is. In fact, if you melt one big cylindrical candle to 5 small cylindrical candles, the sum of the volumes of the smaller candles is equal to the volume of the bigger candle. Hence, when you convert one solid shape to another, all you need to remember is that the volume of the original, as well as the new solid, remains the same.
There are some solid shapes whose volume is given here:
Volume of cylinder = $\pi {r^2}h$
Volume of cube = ${(side)^3}$
Volume of cuboid = $l \times b \times h$
Volume of sphere = $\dfrac{4}{3}\pi {r^3}$
These formulas should be kept in mind for solving these types of questions.
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