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A \( 10 cm \) long wire is brought \( 20 cm \) up in a liquid of surface tension 40 dyne \( cm ^{-1} \). Find the work done against surface tension(A) \( 0.8 m J \)(B) \( 1.6 m J \)(C) \( 1.2 m J \)(D) \( 3.2 mJ \) |
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Answer» \(T = \frac Fl\) F = T x l F = 40 x 10 F = 400 dyne F = 400 x 10-5 N Work done W = F.S = \(400 \times 10^{-5} \times \frac{20}{100}\) = 80 x 10-5 = 0.8 x 10-3 J W = 0.8 mJ |
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