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A thin rectangular magnet suspended free...

A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is `T'`, then the ratio `T'/T` is

A

`1/2`

B

2

C

`1/4`

D

`(1)/(2sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
A

`T=2pi sqrt(1)/(MB),I=(ml^(2))/(I2),M=2 ml `
`T=2pisqrt(I)/(MVB), I=(m)/(2xx12)=1/8M=M//2`
`therefore T./T=1//2`
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