| ⇦ | 
| ⇨ |
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: - In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series
- Equal currents are passing through two very long and straight parallel wires
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- The position vector of a particle R as a function of time is given by
- What determines the nature of the path followed by the particle?
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series
- Equal currents are passing through two very long and straight parallel wires
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- The position vector of a particle R as a function of time is given by
- What determines the nature of the path followed by the particle?
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