A water drop of radius 1cm is broken into 729 equal droplets. If...

The volume of the bigger drop will be, $V = \dfrac{4}{3}\pi{R}^{3}$. If the radius of smaller drops is $r$, using the conservation of volume,

$n\dfrac{4}{3}\pi{r}^{3} = \dfrac{4}{3}\pi{R}^{3} ⇒ 729 \times \dfrac{4}{3}\pi{r}^{3} = \dfrac{4}{3}\pi{R}^{3} ⇒ r = \dfrac{R}{9}$

The initial surface energy will be,

${E}_{i} =$\left(4\pi{R}^{2}\right)T$</p><p>And final surface energy will be,</p><p>${E}_{f} = n\left(4\pi{r}^{2}\right)T = 729 \times 4\pi{\left(\dfrac{R}{9}\right)}^{2}T$$= 36\pi{R}^{2}T$

Therefore, the change in the surface energy will be,

$⇒ \DeltaE = {E}_{f} - {E}_{i} = 32\pi{R}^{2}T = 32 \times 3 . 14 \times {\left(1 \times {10}^{- 2}\right)}^{2} \times 75 \times {10}^{- 3} ⇒ \DeltaE = 7 . 5 \times {10}^{- 4}J