Question.
The diameter of a sphere is decreased by $25 \%$. By what per cent does its curved surface area decrease?
The diameter of a sphere is decreased by $25 \%$. By what per cent does its curved surface area decrease?
Solution:
Let the diameter of the sphere be d.
Radius $\left(r_{1}\right)$ of sphere $=\frac{d}{2}$
New radius $\left(\mathrm{r}_{2}\right)$ of sphere $=\frac{d}{2}\left(1-\frac{25}{100}\right)=\frac{3}{8} d$
$\operatorname{CSA}\left(S_{1}\right)$ of sphere $=4 \pi r_{1}^{2}$
$=4 \pi\left(\frac{d}{2}\right)^{2}=\pi d^{2}$
CSA $\left(S_{2}\right)$ of sphere when radius is decreased $=4 \pi r_{2}{ }^{2}$
$=4 \pi\left(\frac{3 d}{8}\right)^{2}=\frac{9}{16} \pi d^{2}$
Decrease in surface area of sphere $=S_{1}-S_{2}$
$=\pi d^{2}-\frac{9}{16} \pi d^{2}$
$=\frac{7}{16} \pi d^{2}$
Percentage decrease in surface area of sphere $=\frac{S_{1}-S_{2}}{S_{1}} \times 100$
$=\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100=\frac{700}{16}=43.75 \%$
Let the diameter of the sphere be d.
Radius $\left(r_{1}\right)$ of sphere $=\frac{d}{2}$
New radius $\left(\mathrm{r}_{2}\right)$ of sphere $=\frac{d}{2}\left(1-\frac{25}{100}\right)=\frac{3}{8} d$
$\operatorname{CSA}\left(S_{1}\right)$ of sphere $=4 \pi r_{1}^{2}$
$=4 \pi\left(\frac{d}{2}\right)^{2}=\pi d^{2}$
CSA $\left(S_{2}\right)$ of sphere when radius is decreased $=4 \pi r_{2}{ }^{2}$
$=4 \pi\left(\frac{3 d}{8}\right)^{2}=\frac{9}{16} \pi d^{2}$
Decrease in surface area of sphere $=S_{1}-S_{2}$
$=\pi d^{2}-\frac{9}{16} \pi d^{2}$
$=\frac{7}{16} \pi d^{2}$
Percentage decrease in surface area of sphere $=\frac{S_{1}-S_{2}}{S_{1}} \times 100$
$=\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100=\frac{700}{16}=43.75 \%$