A pendulum made of a uniform wire of cross sectional area A has time period 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: (g = gravitational acceleration)

Options

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

Answer

A

Explanation

Time period of simple pendulum,
`T =2 pi sqrt ((l )/(g)) " "…(i)`
If we increase the mass of bob its length is increased by `Delta t.` At that time periodic time is `T_(M)`
`T _(M) =2pi sqrt ((l + Deltal )/(g )) " "…(ii)`
Now Young.s modulus `Y= (F//A)/(Delta l //l ) = (Mg)/(A) xx (1)/( Delta l //l )`
`therefore (Deltal )/(l) = (Mg )/(Ay) " "...(iii)`
Divide equation (ii) by equation (i) and do square of it
`((T _(M))/( T )) ^(2) = (l + Delta l )/(l)= l + (Delta l )/(l)`
`therefore ((T _(M))/(T)) ^(2) -1 = (Delta l )/(l) " "...(iv)`
Putting value of equation (iii) in equation (iv)
`((T_(M))/( T)) ^(2) -1 = (Mg )/(AY) = (A)/( Mg ) [ ((T _(M))/( T )) ^(2) -1] (1)/(Y)`