Mechanical Properties of Fluids - Result Question 17
17. A certain number of spherical drops of a liquid of radius ’ $r$ ’ coalesce to form a single drop of radius ’ $R$ ’ and volume ’ $V$ ‘. If ’ $T$ ’ is the surface tension of the liquid, then :
[2014]
(a) energy $=4 VT(\frac{1}{r}-\frac{1}{R})$ is released
(b) energy $=3 VT(\frac{1}{r}+\frac{1}{R})$ is absorbed
(c) energy $=3 VT(\frac{1}{r}-\frac{1}{R})$ is released
(d) energy is neither released nor absorbed
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Answer:
Correct Answer: 17. (c)
Solution:
- (c) Volume same $\Rightarrow n(\frac{4}{3} \pi r^{3})=\frac{4}{3} \pi R^{3}$
$\Rightarrow n=\frac{R^{3}}{r^{3}}$
Change in energy $=T \Delta A=T[4 \pi R^{2}-n 4 \pi r^{2}]$
$=4 \pi T[R^{2}-\frac{R^{3}}{r^{3}} r^{2}]$
$ =3(\frac{4}{3} \pi R^{3}) T[\frac{1}{R}-\frac{1}{r}] $
$=3 VT\frac{1}{R}-\frac{1}{r}$
$=3 VT[\frac{1}{r}-\frac{1}{R}]$ is released
Energy released when $n$ drops of radius $r$ coalesec to form a body drop of radius $R$,
Energy released $=4 \pi R^{3} T[\frac{1}{r} \frac{-1}{R}]$
If this energy get converted into kinetic energy of big drop, then
$ \begin{aligned} & \frac{1}{2} m v^{2}=4 \pi R^{3} T[\frac{1}{r} \frac{-1}{R}] \\ & \Rightarrow \frac{1}{2}[\frac{4}{3} \pi R^{3} d] V^{2}=4 \pi R^{3} T[\frac{1}{r} \frac{-1}{R}] \\ & \Rightarrow v 2=\frac{6 T}{d}[\frac{1}{r} \frac{-1}{R}] \end{aligned} $
$\Rightarrow$ Velocity of big drop $V=\sqrt{\frac{6 T}{d}(\frac{1}{r} \frac{-1}{R})}$
(Where, $d=$ density of big liquid drop)
