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Question

Determine the energy stored in the surface of a soap bubble of radius 2.1 cm if its surface tension is 4.5×102 N/m.

A
3.6π×104 J
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B
0.6π×104 J
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C
1.6π×104 J
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D
2.6π×104 J
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Solution

The correct option is C 1.6π×104 J
Given,
Radius of soap bubble, R=2.1 cm=2.1×102 m
Surface Tension, T=4.5×102 N/m
We know that, number of free surfaces for soap bubble, n=2
Surface energy = Surface tension × surface area × number of free surfaces
E=T×A×n
=4.5×102×4π(2.1×102)2×2
[ surface area =4πR2 ]
=1.6π×104 J

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