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Question

If T is the surface tension of a liquid, the energy needed to break a liquid drop of radius R into 64 drops is

A
6πR2T
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B
2πR2T
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C
12πR2T
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D
8πR2T
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Solution

The correct option is C 12πR2T
Given,
Radius of larger drop =R
Surface Tension =T
Number of smaller drop, n=64
Let us suppose, radius of smaller drop is r.
As volume of liquid will be same.
Volume of larger drop = Volume of n smaller drops
43πR3=n×43πr3
R3=64×r3
r=R4 .........(1)
We know that, change in energy = surface tension × change in surface area
ΔE=TΔA
=T×4π[64×(R/4)2(R)2)]
[ from (1) and surface area =4πr2 ]
=12πR2T

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