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Question

Calculate the energy released when 1000 small water drops each of same radius 107 m coalesce to form one large drop. The surface tension of water is 5×102 N/m.

A
5π×1012 J
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B
4.2π×1012 J
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C
3.6π×1012 J
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D
1.8π×1012 J
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Solution

The correct option is D 1.8π×1012 J
Let r be the radius of smaller drop and R of the bigger one.
Equating the initial and final volumes, we have
43πR3=1000×43πr3
R=10r=10×107 m
As we know,
Water drops have only one free surface.
Net Change in surface area;ΔA=4πR21000×4πr2
=4π(106)21000×4π(107)2
=36π×1012 m2
Here, negative sign implies that surface area is decreasing. Hence energy released in the process,
W=T×|ΔA|=0.05×36π×1012 m2=1.8π×1012 J

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