A large number of water drops, each of radius r, combine to have … - Vidyadip

MCQ

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
$\frac{2 T }{ rJ }$
C
$\frac{3 T }{ rJ }$
✓
$\frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$

✓

Answer

Correct option: D.

$\frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$

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 $nr ^{3}= R ^{3}$

$\therefore \frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$

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