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Q.

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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a

0.75

b

12

c

1.5

d

6

answer is C.

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Detailed Solution

T=2πIMBH⇒T1T2=B2B1⇒T2=T1B1B2=32=1.5sec

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