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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
- If energy(E), velocity(V) and time(T) are chosen as fundamental quantities,
- Two point charge +9e and +e are at 16 cm away from each other.
- A solenoid 1.5 m long and 0.4 cm in diameter possesses 10 turns per cm length
- A parallel beam of fast moving electrons is incident normally on a narrow slit.
- Two spherical bodies of mass M and 5 M and radii R and 2 R released in free space
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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