Question

If $\sigma$ be the surface tension, the work done in breaking a big drop of radius $R$ in $n$ drops of equal radius is

• A. $R{n}^{2/3}\sigma$
• B. $(n^{2/3}-1)R\sigma$
• C. $(n^{1/3}-1)R\sigma$
• D. $4\pi R^2(n^{1/3}-1)\sigma$
• E. $\frac{1}{n^{1/3}-1}\sigma R$

Answer 1

Answer: (D)

$4\pi R^2(n^{1/3}-1)\sigma$

As we know,

$W=T\Delta A=\sigma (n4\pi r^2-4\pi R^2)$

Again,

$n{r}^3=R^3$ (volume same )

putting it in above

$W=4\pi \sigma (n\times n^{-\frac{2}{3}}{R}^2-R^2)$

hence,

$W=4\pi R^2(n^{1/3}-1)\sigma$