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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?
- A96
- B92
- C86
- D83
Answer not set
Discussion 1
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CCommunity
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%
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%