Properties Of Solids And Liquids Question 343
Question: A mercury drop of 1 cm radius is broken into $ 10^{6} $ small drops. The energy used will be (surface tension of mercury is $ 35\times {{10}^{-3}}N/cm) $ [Roorkee 1984]
Options:
A) $ 4.4\times {{10}^{-3}}J $
B)$ 2.2\times {{10}^{-4}}J $
C) $ 8.8\times {{10}^{-4}}J $
D)$ 10^{4}J $
Show Answer
Answer:
Correct Answer: A
Solution:
$ E=4\pi R^{2}T({{n}^{1/3}}-1) $
$ =4\times 3.14\times {{10}^{-4}}\times 35\times {{10}^{-1}}({{10}^{6/3}}-1) $
$ =4.4\times {{10}^{-3}}J $
