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A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.​

User Randombee
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1 Answer

25 votes
25 votes

Answer:


T=3*10^-3 N/m

Step-by-step explanation:

From the question we are told that:

Radius :


R_1=4=>0.04\\\\R_2=6=>0.06

Work
W=1.5 * 10^(-4)

Generally the equation for Work done is mathematically given by


W=TdA

Where


dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)


dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)


dA=0.050m^2

Therefore


W=TdA


T=(1.5 * 10^(-4))/(0.05m^2)


T=3*10^-3 N/m

User Hzane
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