The radius of a sphere is doubled. Find the increase per cent in its surface area.

Let original radius be r units and new radius be R units.

Given, radius of a sphere is doubled.

∴ R = 2 × r = 2r

Let the original surface area be s and new surface area be S.

By formula,

Percentage increase in surface area = $\dfrac{\text{S - s}}{\text{s}} \times 100$

$= \dfrac{4 π\text{R}^2 - 4 π\text{r}^2}{4π\text{r}^2} \times 100 \\[1em] = \dfrac{4 π(\text{R}^2 - \text{r}^2)}{4π\text{r}^2} \times 100 \\[1em] = \dfrac{(\text{R}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \dfrac{(\text{(2r)}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \dfrac{(4\text{r}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \dfrac{3\text{r}^2}{\text{r}^2} \times 100 \\[1em] = 300 \%.$

Hence, percentage increase in surface area is 300%.