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A liquid soap film in shape of a plane loop has an initial area 0.05 m2. If its area is slowly doubled then theincrease in its surface potential energy from its initial value will be (Surface tension of liquid 0.2 N/m)(A) 5 x 10- J(B) 2 x 102 J(C) 3 x 10-2 J(D) of these |
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Answer» dE = TdsT = surface tension , ds change in area =(0.05)2(0.2) = 2*10-3joules |
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