A pendulumd made of a uniform wire of cross sectional area (A) has time T.When an additionl 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:

Options

A. `[((T_(M))/(T))^(2)-1](A)/(Mg)`
B. `[((T_(M))/(T))^(2) - 1](Mg)/(A)`
C. `[1 - ((T_(M))/(T))-1](A)/(Mg)`
D. `[1- ((T)/(T_(M)))^(2) -1](A)/(Mg)`

Answer

A

Explanation

`T = 2pi sqrt((l)/(g))`
`T_(M) = 2pi sqrt((l + Delta l)/(g)) Delta l = (Mgl)/(AY)`
`(T_(M))/(T) = sqrt((l + Delta l)/(l))`
`((T_(M))/(T))^(2) = 1 + (Mg)/(AY)`
`(1)/(y) = (((T_(M))/(T))^(2) - 1)(A)/(Mg)`