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A bar magnet suspended freely in uniform magnetic field is vibrating with a time period of 3 seconds. If the field strength is increased to four times of the earlier field strength, the time period will be (in seconds)

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answer is 3.

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Detailed Solution

3=2πIMB→(1) T=2πI4MB→(2)12, 3T=41=2, T=32=1.5 sec

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