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.
