Fluid Mechanics Viscosity Question 330
Question: Work done in splitting a drop of water of 1 mm radius into 106 droplets is (Surface tension of water $ =72\times {{10}^{-3}}J/m^{2}) $ [MP PET/PMT 1988; CPMT 1989; RPET 2001]
Options:
A) $ 9.58\times {{10}^{-5}},J $
B) $ 8.95\times {{10}^{-5}},J $
C) $ 5.89\times {{10}^{-5}},J $
D)$ 5.98\times {{10}^{-6}}J $
Show Answer
Answer:
Correct Answer: B
Solution:
Work done in splitting a water drop of radius R into n drops of equal size$ =4\pi R^{2}T({{n}^{1/3}}-1) $ $ =4\pi \times {{({{10}^{-3}})}^{2}}\times 72\times {{10}^{-3}}\times ({{10}^{6/3}}-1) $ $ =4\pi \times {{10}^{-6}}\times 72\times {{10}^{-3}}\times 99=8.95\times {{10}^{-5}}J $
BETA
