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Question
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
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Solution 1
The weight that the soap film supports, W = 1.5 × 10–2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴Total length = 2l = 2 × 0.3 = 0.6 m
Surface tension, `S = "Force of Weight"/"2l"`
`= (1.5 xx 10^(-2))/0.6 = 2.5 xx 10^(-2) "N/m"`
Therefore, the surface tension of the film is 2.5 × 10–2 N m–1.
Solution 2
In present case force of surface tension is balancing the weight of 1.5 x 10-2 N, hence force of surface tension, F = 1.5 x 10-2 N.
Total length of liquid film, l = 2 x 30 cm = 60 cm = 0.6 m because the liquid film has two surfaces.
Surface tension, T = F/l =1.5 x 10-2 N/0.6m =2.5 x 10-2 Nm-1
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