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A pendulum made of a uniform wire of cro...

A pendulum made of a uniform wire of cross sectional area A has time T. When an additional mass M is added to its bob, the time period changes to `T_M`. If the Young's modulus of the material of the wire is Y then `1/Y` is equal to:

A

`[((T_(M))/(T))^(2)-1](A)/(Mg)`

B

`[((T_(M))/(T))^(2)-1](Mg)/(A)`

C

`[-1((T_(M))/(T))^(2)](A)/(Mg)`

D

`[-1((T)/(T_(M)))^(2)](A)/(Mg)`

Text Solution

Verified by Experts

The correct Answer is:
A
`T = pisqrt((l)/(g))`
`T_(M) = 2pisqrt((l + Deltat)/(g)), Deltal = (Mgl)/(AY)`
`T_(M)/(T) = sqrt((l + Deltal)/(l)`
`((T_(M))/(T))^(2) = 1 + (Deltal)/(l), ((T_(M))/(T))^(2) = 1 + (Mg)/(AY)`
`(1)/(y) = (((T_(M))/(T))^(2) - 1 )(A)/(Mg)`
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