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Physics Question on Surface tension

A mercury drop of radius 1 cm1 \,cm is sprayed into 10610^{6} drops of equal size. The energy expended in joule is (surface tension of mercury is 460×10−3 N/m460 \times 10^{-3}\,N / m ):

A

0.0570.057

B

5.75.7

C

5.7×10−45.7 \times 10^{-4}

D

5.7×10−65.7 \times 10^{-6}

Answer

0.0570.057

Explanation

Solution

R=1 cm,n=106R =1 \,cm , n =10^{6} Δ⋃=4πR2(n1/3−1)T\Delta \bigcup=4 \pi R ^{2}\left( n ^{1 / 3}-1\right) T =4π×10−4(102−1)460×10−3=4 \pi \times 10^{-4}\left(10^{2}-1\right) 460 \times 10^{-3} =0.057 J=0.057\, J