Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The length of the magnet is large compared to its width and breadth. The time period of its oscillation in vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts which are then placed on each other with their like poles together. The time period of this combination will be
PhysicsOscillationsNEET
Options:
  • A 23 s
  • B 23  s
  • C 32 s
  • D 32 s
Solution:
1667 Upvotes Verified Answer
The correct answer is: 23 s
The time period of oscillations of the magnet

T=2πIMH

Where, I= moment of inertia of magnet

= ml212 (m, being the mass of magnet)

M=pole strength×L

And H= horizontal component of Earth's magnetic field

When then three equal parts of magnet are placed on one another with their like poles together, then

I=112m3 L32×3=112mL29=I9

And M=Pole strength×L3×3=M 

Hence, T=2πI9MHT=13×T

T=23 s

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.