Question

Let $A$ and $B$ be two solid spheres such that the surface area of $B$ is $300\%$ 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 $k$ must be

Accepted Answer

D. Surface area of sphere $S \propto r^2$. $S_B = S_A + 3 S_A = 4 S_A \implies (r_B / r_A)^2 = 4 \implies r_B / r_A = 2$. Volume of sphere $V \propto r^3$. $V_B / V_A = (r_B / r_A)^3 = 2^3 = 8 \implies V_A = \frac{1}{8} V_B$. Percentage by which $V_A$ is lower than $V_B = \left(1 - \frac{1}{8}ight) 	imes 100\% = \frac{7}{8} 	imes 100\% = 87.5\%$.

Sphere Surface Area and Volume Percentage Difference

Explanation

Practise this under exam conditions