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}) $



NCERT Chapter Video Solution

Dual Pane