Question

A liquid drop having surface energy '$E$' is spread into $512$ droplets of same size. The final surface energy of the droplets is

• A. $2E$
• B. $4E$
• C. $8E$
• D. $12E$

Answer 1

The correct option is C $8E$

Let the surface tension and the density of the liquid be $T$ and $\rho$, respectively.

Radius of bigger drop and small droplets be $a$ and $b$, respectively.

As mass of the liquid is conserved before and after the spreading of drop.

$\therefore$ $\rho \left(\frac{4\pi }{3}{a}^3\right)=\rho (512\times \frac{4\pi }{3}{b}^3)$

Or $a^3=512b^3$

$⟹$ $a=8b$

Surface energy of bigger drop $E=AT=T\left(4\pi a^2\right)$

Total surface energy of small droplets $E^{\prime }=T(512\times 4\pi b^2)$

Or $E^{\prime }=T(512\times 4\pi \frac{a^2}{64})=8T\left(4\pi b^2\right)$

$⟹$ $\frac{E^{\prime }}{E}=\frac{8\times 4\pi T{a}^2}{4\pi T{a}^2}=8$

Thus we get $E^{\prime }=8E$