Question

A liquid drop of diameter D breaks up into 27 drops. Find the resultant change in energy.

• A. $2\pi T{D}^2$
• B. $\pi T{D}^2$
• C. $\frac{\pi T{D}^2}{2}$
• D. $4\pi T{D}^2$

Answer 1

The correct option is A $2\pi T{D}^2$

In this problem volume will be conserved.
So, initial volume is $\frac{4}{3}\pi (\frac{D}{2}{)}^2$
Final volume will be $\frac{4}{3}\pi (r{)}^2\times 27$
So, by volume conservation
$\frac{4}{3}\pi (\frac{D}{2}{)}^2$ = $\frac{4}{3}\pi (r{)}^2$
r = $\frac{D}{6}$
So, initial energy is $4\pi (\frac{D}{2}{)}^2$ T
final energy is $4\pi (\frac{D}{6}{)}^2$ $T\times 27$
Change in energy = final energy - initial energy
= $4\pi (\frac{D}{6}{)}^2\times T\times 27-4\pi (\frac{D}{2}{)}^2\times T$
= $4\pi T{D}^2[\frac{27}{36}-\frac{1}{4}]$
= $4\pi T{D}^2\left[\frac{18}{36}\right]$
= $2\pi T{D}^2$