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Question

A rectangular film of liquid is extended from (4cm×2cm) to (5cm×4cm). If the work done is 3×104J, the value of the surface tension of the liquid is

A
8.0Nm1
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B
0.250Nm1
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C
0.125Nm1
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D
0.2Nm1
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Solution

The correct option is C 0.125Nm1
Initial area of the film Ai=4×2=8 cm2=8×104 m2
Final area of the film Af=5×4=20 cm2=20×104 m2
Thus change in area ΔA=(208)×104=12×104 m2
Work done W=3×104 J
Using for soap films W=S(2ΔA) where S is the surface tension
3×104=S×2×12×104 S=0.125 Nm1

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