| ⇦ | 
| ⇨ |
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: - The velocity of a particle at an instant is 10 ms⁻¹.After 3 s its velocity will become
- A thin circular ring of mass M and radius R rotates about an axis through its centre
- A body weighs 50 grams in air and 40 grams in water. How much would it weigh in a liquid
- The original temperature of a black body is 727⁰C. The temperature at which this black
- If a nucleus ᴢXᴬ emits 9α and 5β-particles, then the ratio of total protons and neutrons
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The velocity of a particle at an instant is 10 ms⁻¹.After 3 s its velocity will become
- A thin circular ring of mass M and radius R rotates about an axis through its centre
- A body weighs 50 grams in air and 40 grams in water. How much would it weigh in a liquid
- The original temperature of a black body is 727⁰C. The temperature at which this black
- If a nucleus ᴢXᴬ emits 9α and 5β-particles, then the ratio of total protons and neutrons
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