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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?1). 962). 923). 864). 83 |
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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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