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Question

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

A
8.0 N/m
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B
0.250 N/m
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C
0.125 N/m
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D
0.2 N/m
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Solution

The correct option is C 0.125 N/m

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)
2ΔA because both side area will be taken (top and bottom)
where S is the surface tension
3×104=S×2×12×104
S=0.125 N/m


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