Question

If the volume of a sphere increases by $72.8\%$, then its surface area increases by

• A. $20\%$
• B. $44\%$
• C. $24,3\%$
• D. $48,6\%$

Answer 1

Answer: (B)

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 $V^{\prime }$ and radius is $r^{\prime }$.
$\therefore V^{\prime }=\frac{4}{3}\pi r^3$
$\Rightarrow V^{\prime }=(V+728\%$ of $V)$
$=V\left(1+\frac{728}{1000}\right)=\frac{1728v}{1000}$
$\Rightarrow V^{\prime }=\frac{1728}{1000}\left(\frac{4}{3}\pi r^3\right)$
$\therefore \frac{1728}{1000}\cdot \frac{4}{3}\pi r^3=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{r^3}{r^3}=1.728$
$\Rightarrow \frac{r^{\prime }}{r}=1.2$
$\therefore$ Increase in surface area of sphere
$={\left(\frac{r^{\prime }}{r}\right)}^2=1.44=144\%$
$\therefore$ Surface area increase $=(144-100)\%$
$=44\%$