Question

A solid metal sphere of surface area $S_1$ is melted and recast into a number of smaller spheres. $S_2$ is the sum of the surface areas of all the smaller spheres. Then,

• A. $S_1$ > $S_2$
• B. $S_2$ > $S_1$
• C. $S_1$ = $S_2$
• D. $S_1$ > $S_2$ only if all the smaller spheres of equal radii

Answer 1

The correct option is B $S_2$ > $S_1$

Let the radius of bigger sphere be $R$ and smaller one be $r_1,r_2,r_3...,r_n$
Thus, volume of bigger sphere = sum of volumes of smaller spheres.
$\frac{4}{3}\pi R^3=\frac{4}{3}\pi ({r_1}^3+{r_2}^3+{r_3}^3...+{r_n}^3)$....(I)
Now, $S_1=4\pi R^2$
$S_2=4\pi (r_1^2+r_2^2+r_3^2+....+r_n^2)$
From (I), we get
$R^3=r_1^3+r_2^3+r_3^3+.....+r_n^3$
If all smaller spheres are of equal radius:
$R^3=n{r}^3$
Now, $\frac{S_1}{S_2}=\frac{R^2}{n{r}^2}=\frac{(n{r}^3{)}^{\frac{2}{3}}}{n{r}^2}$ = $\frac{n^{\frac{2}{3}}}{n}$
$\Rightarrow n{S}_1=n^{\frac{2}{3}}{S}_2$
$\Rightarrow n^{\frac{1}{3}}{S}_1=S_2$
Thus, $S_2>S_1$
Hence, option 'B' is correct.