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Question

Two drops of equal radius coalesce to form a bigger drop. What is the ratio of surface energy of the bigger drop to the smaller drop?

A
21/3:1
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B
1:1
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C
22/3:1
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D
1:2
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Solution

The correct option is C 22/3:1
Let us suppose the surface tension is T, the radius of smaller drop is r and that of bigger drop is R.
As total volume of fluid will be same
Volume of two smaller drops = Volume of bigger drop
2×43πr3=43πR3
R=21/3r ......(1)
Now, ratio of surface energies
EbigEsmall=T×4πR2T×4πr2
[ surface energy =surface tension × surface area ]
=R2r2=22/3:1 [ from (1) ]

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