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Question

Physics Question on Surface tension

A mercury drop of radius 10−3m10^{-3} m is broken into 125125 equal size droplets Surface tension of mercury is 045 Nm−1045\, Nm ^{-1} The gain in surface energy is:

A

5×10−5J5 \times 10^{-5} J

B

17.5×10−5J17.5 \times 10^{-5} J

C

2.26×10−5J2.26 \times 10^{-5} J

D

28×10−5J28 \times 10^{-5} J

Answer

2.26×10−5J2.26 \times 10^{-5} J

Explanation

Solution

Initial surface energy =0.45×4π(10−3)2
34​π(10−3)3=125×34π​Rnew 3​
∴10−3=5Rnew ​
∴Rnew ​=510−3​m
So, final surface energy =0.45×125×4π(510−3​)2
Increase in energy =0.45×4π×(10−3)2[25125​−1] =4×0.45×4π×10−6
=2.26×10−5J