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Q.
If the volume of a sphere increases by $72.8 \%$, then its surface area increases by
KVPYKVPY 2009
Solution:
Let initial volume of sphere is $V$ and radius is $r$.
$\therefore V=\frac{4}{3} \pi r^{3}$
Volume of sphere after increase is $V'$ and radius is $r'$.
$\therefore V'=\frac{4}{3} \pi r^{3}$
$\Rightarrow V'=(V+728 \%$ of $V)$
$=V\left(1+\frac{728}{1000}\right)=\frac{1728 v}{1000}$
$\Rightarrow V'=\frac{1728}{1000}\left(\frac{4}{3} \pi r^{3}\right)$
$\therefore \frac{1728}{1000} \cdot \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi r^{3}$
$\Rightarrow \frac{r^{3}}{r^{3}}=1.728$
$\Rightarrow \frac{r'}{r}=1.2$
$\therefore$ Increase in surface area of sphere
$=\left(\frac{r'}{r}\right)^{2}=1.44=144 \%$
$\therefore$ Surface area increase $=(144-100) \%$
$=44 \%$