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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
- A beam of cathode rays is subjected to crossed Electric (E) and Magnetic fields (B)
- Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
- A metal rod of length l cuts across a uniform magnetic field B with a velocity v.
- A parallel plate air capacitor has capacity C, distance of seperation between plate
- The distance of a geo-stationary satellite from the centre of the earth is nearest to
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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