Question

A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J ,$ the rise in heat energy per unit volume will be:

• A. $\frac{2 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$
• B. $\frac{2T }{ rJ }$
• C. $\frac{3 T }{ rJ }$
• D. $\frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$

Answer 1

Answer: (D)

$n \times \frac{4}{3} \pi r ^{3}=\frac{4}{3} \pi R ^{3}$
$\therefore n ^{1 / 3} r = R$
$\therefore $ Total change in surface energy
$=\left( n \left(4 \pi r ^{2}\right)-4 \pi R ^{2}\right) T$
$\Rightarrow 4 \pi T \left( nr ^{2}- R ^{2}\right)$
$\therefore $ Heat energy
$=\frac{4 \pi T \left( nr ^{2}- R ^{2}\right)}{ J \times \frac{4}{3} \pi R ^{3}}=\frac{3 T }{ J }\left(\frac{ nr ^{2}}{ R ^{3}}-\frac{1}{ R }\right)$
Put $nr ^{3}= R ^{3}$
$\therefore \frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$