Question

14. If the radius of a sphere is increased by 20 %, by what per cent does its volume
increase?​

Answer 1

Step-by-step explanation:

$let \: the \: original \: radius \: be \: r \: and \: \\ volume \: be \: v \: then \\ v = \frac{4}{3} \pi {r}^{3} \\ now \: new \: radius =original \: radius \: + 20\% \: of \: r \\ = r + \frac{20r}{100} = \frac{6r}{5} \\ therefore \: new \: volume \: = \frac{4}{3} \pi{(\frac{6r}{5})}^{3} \\ = \frac{4}{3} \pi {r}^{3} \times \frac{216}{125} \\volume \: increased = \frac{4}{3} \pi {r}^{3} \times \frac{216}{125} \: - \frac{4}{3} \pi {r}^{3} = \frac{4}{3} \pi {r}^{3}( \frac{91}{125} ) \\ \% \: of \: volume \: inreased = \frac{increased \: volume}{original \: volume} \times 100\% \\ = \frac{\frac{4}{3} \pi {r}^{3}( \frac{91}{125} )}{ \frac{4}{3} \pi {r}^{3} } \times 100 \\ = \frac{9100}{125} = 72.8\%$