| ⇦ | 
| ⇨ |
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: - When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- Two identical capacitors each of capacitance 5 mF are charged potentials 2 kV and 1 kV
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- Frequency of stretched string will increase by
- If two electric bulbs have 40 W and 60 W ratings at 220 V, then the ratio
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- Two identical capacitors each of capacitance 5 mF are charged potentials 2 kV and 1 kV
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- Frequency of stretched string will increase by
- If two electric bulbs have 40 W and 60 W ratings at 220 V, then the ratio
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