Question

The radius of the sphere increased by $10\%.$ What is the percentage increase in the volume of the sphere?

Answer 1

We know that the volume of a sphere with radius $“r”$ is given by the equation:
Volume of the sphere $=\frac{4}{3}\pi r^3$ …………….. eq (1)
$10\%$ increase in radius: $r\to r+10\%r$
$\therefore$ Increased radius $=r+\frac{1}{10}r=\frac{11}{10}r$
The volume of the new sphere $=\frac{4}{3}\pi {\left(\frac{11}{10}r\right)}^3$ (substituting for the increased r)
$=\frac{4}{3}\pi \frac{1331}{1000}{r}^3$
$=\frac{4}{3}\pi \times 1.331r^3$……………… eq (2)
Increase in volume = Volume of the new sphere − Volume of the original sphere
= Equation (2) − Equation (1)
$=\frac{4}{3}\pi \times 1.331r^3-\frac{4}{3}\pi r^3$
$=\frac{4}{3}\pi r^3\times (1.331-1)$
$=\frac{4}{3}\pi r^3\times 0.331$
Percentage increase in the volume $=\frac{\text{Increased volume of the sphere}}{\text{Original volume of the sphere}}\times 100$
$=\frac{\frac{4}{3}\pi r^3\times 0.331}{\frac{4}{3}\pi r^3}\times 100$
$=33.1\%$

Hence, the percentage increase in the volume of the sphere is $33.1\%.$
$\to$ Option D is correct.