Properties Of Solids And Liquids Question 360
Question: A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is $ 0.465\ J/m^{2} $ )[UPSEAT 2002]
Options:
A) $ 23.4\mu J $
B) $ 18.5\mu J $
C) $ 26.8\mu J $
D)$ 16.8\mu J $
Show Answer
Answer:
Correct Answer: A
Solution:
Increase in surface energy or work done in splitting a big drop $ =4\pi R^{2}T({{n}^{1/3}}-1) $
$ \Rightarrow W=4\pi \times {{(2\times {{10}^{-3}})}^{2}}\times 0.465({{8}^{1/3}}-1)=23.4\ \mu J $
BETA
