| ⇦ | 
| ⇨ |
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be:
Options
(a) 2 √3 J
(b) 3 J
(c) √3 J
(d) 3/2 J
Correct Answer:
3 J
Explanation:
According to work energy theorem
W = U final – U initial = MB (cos 0 – cos 60⁰)
W = MB/2 = √3J …(i)
? = M⃗ x B⃗ = MB sin 60⁰ = (MB √3 / 2) …(ii)
From eq (i) and (ii)
? = 2 √3 x √3 / 2 = 3 J
Related Questions: - A spring balance is attached to the ceiling of a lift.A man hangs his bag on the spring
- The work done in which of the following processes is equal to the change in internal energy
- A piece of iron is heated in a flame. It first comes dull red then becomes reddish
- A particle is executing a simple harmonic motion. Its maximum acceleration
- A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A spring balance is attached to the ceiling of a lift.A man hangs his bag on the spring
- The work done in which of the following processes is equal to the change in internal energy
- A piece of iron is heated in a flame. It first comes dull red then becomes reddish
- A particle is executing a simple harmonic motion. Its maximum acceleration
- A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature
Topics: Magnetic Effects of Current and Magnetism (167)
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