Properties Of Solids And Liquids Question 328
Question: Energy needed in breaking a drop of radius R into n drops of radii r is given by [CPMT 1982, 97]
Options:
A) $ 4\pi T(nr^{2}-R^{2}) $
B)$ \frac{4}{3}\pi (r^{3}n-R^{2}) $
C) $ 4\pi T(R^{2}-nr^{2}) $
D)$ 4\pi T(nr^{2}+R^{2}) $
Show Answer
Answer:
Correct Answer: A
Solution:
Energy needed = Increment in surface energy = (surface energy of n small drops) - (surface energy of one big drop) $ =n4\pi r^{2}T-4\pi R^{2}T=4\pi T(nr^{2}-R^{2}) $
