MCQ
A liquid drop having surface energy $E$ is spread into 512 droplets of same size. The final surface energy of the droplets is
  • A
    $2 E$
  • B
    $4 E$
  • $8 E$
  • D
    $12 E$

Answer

Correct option: C.
$8 E$
(c) : Let $R$ and $r$ be the radius of big drop and each droplet respectively.
$
\therefore \quad \frac{4 \pi R^3}{3}=512 \times \frac{4 \pi}{3} r^3 \Rightarrow R=8 r
$
Now, surface energy of big drop is
$
E=4 \pi R^2 T
$
Total surface energy of small droplets is
$
\begin{aligned}
E_1 & =512 \times 4 \pi r^2 T \\
& =512 \times 4 \pi\left(\frac{R}{8}\right)^2 T=8 \times 4 \pi R^2 T=8 E
\end{aligned}
$

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