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Question

The surface tension of soap solution is T=0.03 Nm1. If the size of a soap bubble is increased from a radius of 3 cm to 5 cm, the work done in increasing the size of the soap bubble is

A
4π mJ
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B
0.2π mJ
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C
2π mJ
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D
0.4π mJ
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Solution

The correct option is D 0.4π mJ
Given,
Surface tension of soap solution (T)=0.03 Nm1
Initial radius of soap bubble (r1)=3 cm
Final radius of soap bubble (r2)=5 cm
Soap bubble has two surface areas
Hence,
W=TΔA=T×[2×4π(r22r21)]
0.03[2×4π(5232)]×104
=0.384π×103 J0.4π mJ
Option (d) is the correct answer

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