MCQ
$8$ mercury drops coalesce to form one mercury drop, the energy changes by a factor of
- A$1$
- B$2$
- ✓$4$
- D$6$
✓
Answer
Correct option: C.
$4$
c
(c) As volume remains constant therefore $R = {n^{1/3}}r$
(c) As volume remains constant therefore $R = {n^{1/3}}r$
$\frac{{{\rm{Energy\, of \,big \,drop}}}}{{{\rm{Energy \,of\, small\, drop}}}} = \frac{{4\pi {R^2}T}}{{4\pi {r^2}T}} = \frac{{{R^2}}}{{{r^2}}}$$ = {(8)^{2/3}} = 4$
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