Question

$8$ drops of equal radius coleasec to form a bigger drop. The ratio of surface energy of bigger drop to smaller one is?

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is D $4:1$

Let r$=$ radius of each small drop
R$=$ radius of bigger drop
As $8$ small drops colease to form a bigger drop.
So, $8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\therefore R=(8{)}^{1/3}r=2r$
Surface energy $\propto$ surface area
So, $\frac{E_b}{E_s}=\frac{4\pi R^2}{4\pi r^2}={\left(\frac{R}{r}\right)}^2={\left(\frac{2r}{r}\right)}^2=4$
$\therefore \frac{E_b}{E_s}=4:1$.