| ⇦ | 
| ⇨ |
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.
Related Questions: - According to Hook’s law of elasticity, if stress is increased, the ratio of stress
- Length of a conductor is 50 cm radius of cross section 0.1 cm and resistivity
- The force of repulsion between two electrons at a certain distance is F.
- A tuning fork frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second
- Energy bands in solids are a consequence of
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- According to Hook’s law of elasticity, if stress is increased, the ratio of stress
- Length of a conductor is 50 cm radius of cross section 0.1 cm and resistivity
- The force of repulsion between two electrons at a certain distance is F.
- A tuning fork frequency 512 Hz is vibrated with a sonometer wire and 6 beats per second
- Energy bands in solids are a consequence of
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply