If the radius of a sphere is increased by 50%, find the increase per cent in its volume.

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

Given, radius of a sphere is increased by 50%.

∴ R = r + $\dfrac{50}{100}$ × r = r + $\dfrac{1}{2}$ r = r + 0.5 r = 1.5 r

Let the original volume be v and new volume be V.

By formula,

Percentage increase in volume = $\dfrac{\text{V - v}}{\text{v}} \times 100$

$= \dfrac{\dfrac{4}{3} π\text{R}^3 - \dfrac{4}{3} π\text{r}^3}{\dfrac{4}{3} π\text{r}^3} \times 100 \\[1em] = \dfrac{\dfrac{4}{3} π(\text{R}^3 - \text{r}^3)}{\dfrac{4}{3} π\text{r}^3} \times 100 \\[1em] = \dfrac{(\text{R}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \dfrac{(\text{(1.5r)}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \dfrac{(3.375\text{r}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \dfrac{2.375\text{r}^3}{\text{r}^3} \times 100 \\[1em] = 237.5 \%$

Hence, percentage increase in volume is 237.5%.