How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm. Surface tension of soap solution is `3.0xx10^(-1) Nm^(-1)`.
Text Solution
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Soap bubble has two surfaces. Hence, `W=T Delta A` Here, `DeltaA=2[4pi{(2.5xx10^(-2))^(2)-(1.0xx10^(-2))^(2)}]` `=1.32xx10^(-2)m^(2)` `therefore " Work done" W=(3.0xx10^(-2))(1.32xx10^(-2))J=3.96xx10^(-4) J`
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