Properties of matterMedium
Question
Work of 3.0 × 10-4 joule is required to be done in increasing the size of a soap film from 10cm × 6cm to 10cm × 11cm. The surface tension of the film is
Options
A.5 × 10-2N/m
B.3 × 10-2N/m
C.1.5 × 10-2N/m
D.1.2 × 10-2N/m
Solution
Area increased = (10 × 11) - (10 × 6)cm2
= 110 - 60 = 50 cm2
∴ total increased area = 50 × 2 = 100 cm2
work done =surface tension × increase in surface area
⇒ Surface tension
= 0.03N/m = 3 × 10-2 N/m
= 110 - 60 = 50 cm2
∴ total increased area = 50 × 2 = 100 cm2
work done =surface tension × increase in surface area
⇒ Surface tension
= 0.03N/m = 3 × 10-2 N/m
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