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 :

  • Option 1)

    \text{energy}=4\text{VT}\left ( \frac{1}{\text{r}} -\frac{1}{\text{R}}\right )\ \text{is}\ \text{released}.

  • Option 2)

    \text{energy}=3\text{VT}\left ( \frac{1}{\text{r}} +\frac{1}{\text{R}}\right )\ \text{is}\ \text{absorbed}.

  • Option 3)

    \text{energy}=3\text{VT}\left ( \frac{1}{\text{r}} -\frac{1}{\text{R}}\right )\ \text{is}\ \text{released}.

  • Option 4)

    energy is neither released nor absorbed.

 

Answers (1)

As we discussed in

Since initial volume = final volume

Let's say number of drops = N

Then energy released = T.\:\Delta A

A_{i}\:=\:(4\pi r^{2}).n

n = number of drops

Total volume = n. (\frac{4\pi}{3}r^{3})\:=\:(n.\:4\pi r^{2})\:.\:\frac{r}{3}

Or A_{i}=(n.\:4\pi r^{2})=\frac{3V}{r}

A_{f}=4\pi R^{2}\:=\:\frac{3V}{R}

\therefore Energy released = T.(\frac{3V}{R}-\frac{3V}{r})

=\:3VT\:(\frac{1}{r}-\frac{1}{R})

 


Option 1)

\text{energy}=4\text{VT}\left ( \frac{1}{\text{r}} -\frac{1}{\text{R}}\right )\ \text{is}\ \text{released}.

This option is incorrect.

Option 2)

\text{energy}=3\text{VT}\left ( \frac{1}{\text{r}} +\frac{1}{\text{R}}\right )\ \text{is}\ \text{absorbed}.

This option is incorrect.

Option 3)

\text{energy}=3\text{VT}\left ( \frac{1}{\text{r}} -\frac{1}{\text{R}}\right )\ \text{is}\ \text{released}.

This option is correct.

Option 4)

energy is neither released nor absorbed.

This option is incorrect.

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