Question

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

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

Answer 1

The correct option is C

$\left(2\mathrm{P}+\frac{{\mathrm{P}}^2}{100}\right)\%$

Step 1: Given data

The radius of sphere is increased by $P\%$

The surface area of the sphere will increase by $?\%$

Let, $r$ be the radius of the sphere

When the radius of the sphere is increased by$P\%$,

New radius$=\left(r+\frac{P}{100}\times r\right)=r\left(1+\frac{P}{100}\right)$

Since the Original surface area$=4{\mathrm{\pi r}}^2$

New surface area$=4\mathrm{\pi }{\left[\mathrm{r}\left(1+\frac{\mathrm{P}}{100}\right)\right]}^2=4{\mathrm{\pi r}}^2{\left(1+\frac{\mathrm{P}}{100}\right)}^2$

Step 2: Calculating the increase in the percentage of the surface area of a sphere

An increase in the percentage of the surface area is given by,

$\frac{changeinsurfacearea}{originalsurfacearea}\times 100$

Thus,

$$\begin{gathered} \Rightarrow \frac{4{\mathrm{\pi r}}^2{\left(1+\frac{\mathrm{P}}{100}\right)}^2-4{\mathrm{\pi r}}^2}{4{\mathrm{\pi r}}^2}\times 100 \\ \Rightarrow \frac{4{\mathrm{\pi r}}^2\left[{\left(1+\frac{\mathrm{P}}{100}\right)}^2-1\right]}{4{\mathrm{\pi r}}^2}\times 100 \\ \Rightarrow \left[{\left(1+\frac{\mathrm{P}}{100}\right)}^2-1\right]\times 100 \\ \Rightarrow \left[1^2+\frac{{\mathrm{P}}^2}{{100}^2}+2\times 1\times \frac{P}{100}-1\right]\times 100 \\ \Rightarrow \left[\frac{{\mathrm{P}}^2}{{100}^2}+\frac{2P}{100}\right]\times 100 \\ \Rightarrow \left(2P+\frac{{\mathrm{P}}^2}{100}\right)\% \end{gathered}$$

Hence, the increase in the percentage of the surface area the of the sphere will be $\left(2\mathrm{P}+\frac{{\mathrm{P}}^2}{100}\right)\%$

Hence, option C is correct.