| ⇦ | 
| ⇨ |
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth’s horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be
Options
(a) 1 s
(b) 2 s
(c) 3 s
(d) 4 s
Correct Answer:
4 s
Explanation:
Time period of a vibration magnetometer,
T 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁ √(B₁ / B₂)
2 √ (24 x 10⁻⁶ / 6 x 10⁻⁶) = 4 sec.
Related Questions: - Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
- If the focal length of objective lens is increased, then magnifying power of
- If A=3i+4j and B=7i+24j the vector having the same magnitude as B and parallel to A is
- Van der waals,equation of state is(p+a/v²)(V-b)=nRT. The dimensions of a and b
- Pick out the statement ehich is incorrect?
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
- If the focal length of objective lens is increased, then magnifying power of
- If A=3i+4j and B=7i+24j the vector having the same magnitude as B and parallel to A is
- Van der waals,equation of state is(p+a/v²)(V-b)=nRT. The dimensions of a and b
- Pick out the statement ehich is incorrect?
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