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Question

Energy released when 1000 small water drops each of same radius 107 m coalesce to form one large drop
(Surface tension of water is 7×102 N/m)

A
2π×1012 J
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B
2.5π×1012 J
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C
1.5π×1012 J
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D
5π×1012 J
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Solution

The correct option is B 2.5π×1012 J
Given,
Number of water drops, n=1000
Radius of water drop,r=107 m
Total volume =n×43πr3
=1000×43π×(107)3 ... (1)
Volume of large drop if its radius is R
=43πR3 ... (2)
On equating (1) and (2) as volume will remain same,
43πR3=1000×43π×(107)3
R=106 m
We know, surface energy, U=TΔA
U=TΔA=T(4πR2n×4πr2)
U=7×102×4×π×[(106)21000×(107)2]
U=2.5π×1012 J
(-ve sign indicates release of energy)

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