Question

27 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

• A. 6 times
• B. 10 times
• C. 9 times
• D. 2 times

Answer 1

Answer: (C)

$R=n^{1/3}r$
$\Rightarrow R=3r$
$\frac{\text{ Energy of big drop }}{\text{ Energy of small drop }}=\frac{T\times 4\pi R^2}{T\times 4\pi r^2}$
$\frac{R^2}{r^2}=\frac{(3r{)}^2}{r^2}=9$