Question

The radius of a sphere is increased by $50\%$, then the increase in surface area of a sphere is:

• A. $200\%$
• B. $150\%$
• C. $125\%$
• D. $50\%$

Answer 1

The correct option is C $125\%$

Let the radius of sphere is $r$.

Then surface area $=4\pi r^2$.

If radius increased by $50\%$ new radius $=r\times \frac{150}{100}=\frac{3r}{2}$.

So surface area of new sphere $=4\pi (\frac{3r}{2}{)}^2=4\times \frac{9}{4}\pi r^2=9\pi r^2$.

Then increased in surface area of sphere $=9\pi r^2-4\pi r^2=5\pi r^2$.

So $\%$ increased in $S.A=\frac{5\pi r^2}{4\pi r^2}=\frac{5}{4}=\frac{5}{4}\times \frac{100}{100}=\frac{5\times 25}{100}=125\%$.

Therefore, option $C$ is correct.