Fluid Mechanics Viscosity Question 354
Question: 8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of [DCE 2000]
Options:
A) 1
B) 2
C) 4
D) 6
Show Answer
Answer:
Correct Answer: C
Solution:
As volume remains constant therefore $ R={{n}^{1/3}}r $ $ \frac{\text{Energy of big drop}}{\text{Energy of small drop}}=\frac{4\pi R^{2}T}{4\pi r^{2}T}=\frac{R^{2}}{r^{2}} $ $ ={{(8)}^{2/3}}=4 $
BETA
