Question

The diameter of a sphere is decreased by $25\%$ . By what per cent does its curved surface area decrease?

Answer 1

let the diameter of sphere be 'd'

Radius of sphere $r_1=\frac{d}{2}$

radius of outer sphere $r_2=\frac{d}{2}(1-\frac{25}{100})$

$S_1=4\pi r_1^2$

$\therefore 4\pi (\frac{d}{2}{)}^2$

$S_1=\pi d^2$

the diameter of sphere is decreased by $25$ % then its outer surface area ,

$S_2=4\pi r_2^2$

$=4\pi (\frac{3d}{8}{)}^2$

$S_2=\frac{9}{16}\pi d^2$

Difference of outer surface of spheres

$=S_1-S_2$

$=\pi d^2-\frac{9}{16}\pi d^2$

$=\frac{7}{16}\pi d^2$

% difference of outer surface of sphere

$=\frac{S_2-S_2}{S_1}\times 100$

$=\frac{7\pi d^2}{16\pi d^2}\times 100$

$=\frac{700}{16}$

$=43.75$ %