1.

A thin rectangularmagnet suspeded freeely hasa period of oscilltion equal to T now it is broken into tow equalhalveseach having of original lengthfieldif its period of oscillation is T then(T)/(T)is

Answer»

`1/2`
2
`1/4`
`(1)/(2sqrt(2))`

Solution :`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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