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Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is
Options
(a) 5 x10⁻² N/m
(b) 3 x10⁻² N/m
(c) 1.5 x10⁻² N/m
(d) 1.2 x10⁻² N/m
Correct Answer:
3 x10⁻² N/m
Explanation:
Surface tension, S = work done/increase in area
As the soap film has two surfaces.
S = (3.0 x 10⁻⁴ J)/[2 x (10 x 11 – 10 x 6 ) x 10⁻⁴ m²]
= 3 x 10⁻² N/m.
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Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In stretching a spring by 2 cm, energy stored is given by U. Had is been stretched
- The binding energy of deutron is 2.2 MeV and that of ₂⁴He is 28 MeV.
- A point charge q is situated at a distance r on axis from one end of a thin
- MA⁻¹T⁻² is the dimensional formula of
- In which of the processes, does the internal energy of the system remain constant?
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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