| ⇦ | 
| ⇨ |
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: - The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- An artificial satellite is revolving round the earth in a circular orbit. Its velocity
- The work function of metals is in the range of 2 eV to 5 eV.
- The second overtone of an open pipe is in resonance with the first overtone
- An Electric charge does not have which of the following properties?
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- An artificial satellite is revolving round the earth in a circular orbit. Its velocity
- The work function of metals is in the range of 2 eV to 5 eV.
- The second overtone of an open pipe is in resonance with the first overtone
- An Electric charge does not have which of the following properties?
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