1.

Let A and B be two solid spheres such that the surface area of B is 800% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k is approximately?1). 962). 923). 864). 83

Answer»

We know that surface area of a sphere = 4πr2

Where r = radius

Let the surface area of sphere A = x

But given that surface area of sphere B = x + 800% of surface area of sphere A = x + 800% of x = 9x

⇒ Surface area of sphere A / surface area of sphere B = 4πr2/4πR2 = x/9x

⇒ r/R = 1/√ 9

⇒ R = 3r

We know that VOLUME of sphere = (4/3)πr3

⇒ Volume of sphere A is found to be K% LOWER than the volume of B = (4/3πR3 - 4/3πr3)/4/3πR3 × 100

 = (4/3π(3r)3 - 4/3πr3) / 4/3π(3r)3 × 100 = 96%


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