Question

The radius of a sphere is increased by $P\%$. Its surface area increases by

• A. $P\%$
• B. $P^2\%$
• C. $(2p+\frac{p^2}{100})\%$
• D. $\frac{p^2}{2}\%$

Answer 1

The correct option is C $(2p+\frac{p^2}{100})\%$

Let r be radius of the sphere
New radius $=(r+\frac{p}{100}r)=(1+\frac{p}{100}r)$
Original surface area of sphere $=4\pi r^2$
New surface area
$=4\pi {\left((1+\frac{p}{100})r\right)}^2=4\pi {(1+\frac{p}{100})}^2{r}^2$
$\therefore \%$ increase in surface area
$=\frac{Changeinsurfacearea}{Originalsurfacearea}\times 100$
$(1+\frac{p^2}{(100{)}^2}+\frac{2p}{100}-1)\times 100=(\frac{p^2}{100}+2P)\%$