| ⇦ |
| ⇨ |
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: - A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- Two strings A and B, made of same material are stretched by same tension
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
- When a charged oil drop moves upwards in an electric field,
- The ratio of the radii or gyration of a circular disc to that of a circular ring
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- Two strings A and B, made of same material are stretched by same tension
- A nucleus ᴢXᴬ emits an α- particle with velocity v. The recoil speed of the daughter
- When a charged oil drop moves upwards in an electric field,
- The ratio of the radii or gyration of a circular disc to that of a circular ring
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