### NEET Solved Paper 2014 Question 14

##### Question: 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 [AIPMT 2014]

#### Options:

A) Energy = $ 4VT( \frac{1}{r}-\frac{1}{R} ) $ is released

B) Energy = $ 3VT( \frac{1}{r}+\frac{1}{R} ) $ absorbed

C) Energy = -$ 3VT( \frac{1}{r}-\frac{1}{R} ) $ is released

D) Energy is neither released nor absorbed

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Here, if the surface area changes, it will change the surface energy as well.

If the surface area decreases, it means that energy is released and vice-versa.

Change in surface energy $ \Delta A\times T,……(i) $

Let we have ’n’ number of drops initially. So, $ \Delta A=4\pi R^{2}-n(4\pi r^{2}),……(ii) $ Volume is constant So,

$ n=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}=V……(iii) $

From Eqs. (ii) and (iii) $ \Delta A=\frac{3}{R}\frac{4\pi }{3}\times R^{3}-\frac{3}{r}( n\frac{4\pi }{3r^{3}} ) $ $ =\frac{3}{R}\times V-\frac{3}{r}V $ $ \Delta A=3V( \frac{1}{R}-\frac{1}{r} )=-ve $ value.

As R > r, so $ \Delta A $ is negative. It means surface area is decreased, so energy must be released.

Energy released = $ \Delta A\times T=-3VT( \frac{1}{r}-\frac{1}{R} ) $ Above expression shows the magnitude of energy released.