MCQ
Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $ r$ is given by
- ✓$4\pi T(n{r^2} - {R^2})$
- B$\frac{4}{3}\pi ({r^3}n - {R^2})$
- C$4\pi T({R^2} - n{r^2})$
- D$4\pi T(n{r^2} + {R^2})$
✓
Answer
Correct option: A.
$4\pi T(n{r^2} - {R^2})$
a
(a) Energy needed = Increment in surface energy
(a) Energy needed = Increment in surface energy
= (surface energy of $n$ small drops) -(surface energy of
one big drop)
$ = n4\pi {r^2}T - 4\pi {R^2}T = 4\pi T(n{r^2} - {R^2})$
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