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

Answer

A

Explanation

We know that time period , `T=2pisqrt((L)/(g))`
When additional mass M is added to its bob `T_(M)=2pisqrt((L+DeltaL)/(g))`
where , `DeltaL` is inccrease in length.
We know that `Y=(Mg//A)/(DeltaL//L)=(MgL)/(AdDeltaL)`
`rArrDeltaL=(MgL)/(AY)rArrT_(M)=2pisqrt((L+(MgL)/(AY))/(g))rArr((T_(M))/(T))^(2)=1+(Mg)/(AY)`
or `(Mg)/(AY)=((T_(M))/(T))^(2)-1 or(1)/(Y)=(A)/(mg)[((T_(M))/(T))^(2)-1]`