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The work done in increasing the size of a soap film from `10cmxx6cm` to `10cmxx11cm` is `3xx10^-4` Joule. The surface tension of the film isA. `1.5xx10^(-2)Nm^(-1)`B. `3.0xx10^(-2) Nm^(-1)`C. `6.0xx10^(-2)Nm^(-1)`D. `11.0xx10^(-2) Nm^(-1)` |
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Answer» Correct Answer - B `A_(1) = 10xx6 = 60cm^(2) = 60xx10^(-4)m^(2), A_(2) = 10xx11 = 110 cm^(2) = 110xx10^(-4) m^(2)` As the soap film has two free surfaces `therefore W = T xx2Delta A` `implies W = Txx2xx(A_(2)-A_(1)) implies T (W)/(2xx50xx10^(-4)) =(3xx10^(-4))/(2xx50xx10^(-4))= (3xx10^(-2)) N//m` |
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