| ⇦ | 
| ⇨ |
Radius of a soap bubble is increased from R to 2R. Work done in this process in terms of surface tension S is
Options
(a) 24 πR²S
(b) 48 πR²S
(c) 12 πR²S
(d) 36 πR²S
Correct Answer:
24 πR²S
Explanation:
Work done = 2 x S x Change in surface area
W = 2 x S x 4π (R₂² – R₁²)
Given: R₂ = 2R, R₁ = R = 8S π (4R² – R²) = 24 πR²S.
Related Questions:
- Two bodies A(of mass 1kg) and B(of mass 3 kg) are dropped from heights of 16m and 25m
- If vectors A=cos wti + sin wtj and B= cos wt/2 i + sin wt/2 j are functions of time,
- A spherical body of emissivity e=0.6, placed inside a perfectly black body is maintained
- The drift velocity of the electrons in a copper wire of length 2 m under
- The angular momentum of an electron is J. The magnitude of the magnetic moment
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