Energy needed in breaking a drop of radius R | vedclass

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(A) The energy required to break a large drop into smaller drops is equal to the increase in the total surface energy of the system.$1$. Initial surfa

Vedclass · Accepted Answer

$4\pi T(n{r^2} - {R^2})$

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(A) The energy required to break a large drop into smaller drops is equal to the increase in the total surface energy of the system.$1$. Initial surface energy of the large drop of radius $R$: $U_i = 4\pi R^2 T$,where $T$ is the surface tension.$2$. Final surface energy of $n$ small drops of radius $r$: $U_f = n(4\pi r^2 T) = 4\pi n r^2 T$.$3$. The energy needed is the change in surface energy: $\Delta U = U_f - U_i$.$4$. Therefore,$\Delta U = 4\pi n r^2 T - 4\pi R^2 T = 4\pi T(n r^2 - R^2)$.

Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $r$ is given by

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