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Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of soap solution is 30 dyne/cm. `(pi=3.142)` |
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Answer» Given : Initial radius of soap bubble `(r_(1))=1 cm` `=1xx10^(-2)m` Final radius of soap bubble `(r_(2))` `=2cm=2xx10^(-2)m` Surface tension (T) `=30 ` dyne/cm `=(30xx10^(-5)N)/(10^(-2)m)=30xx10^(-3)N//m` We know that, Work done = T.dA `=30xx 10^(-3)xx4pi xx [2^(2)-1^(2)]xx10^(-4)xx2` `=2.262 xx 10^(-4)` joule |
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