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

Answer

A

Explanation

`T=2pi sqrt((l)/(g))` ….(i),
`T_(M)=2pi sqrt((l+Delta l)/(g))` …..(ii)
`Y=(Fl)/(A Delta l)` or `Delta l=(Fl)/(AY)=(mgl)/(AY)` or `(Delta l)/(l)=(mg)/(AY)`
Dividing (ii) by (i)
`(T_(M))/(T)=sqrt(1+(Delta l)/(l))`, squaring both side we get
`(Delta l)/(l)=[((T_(M))/(T))^(2) -1] (A)/(mg)`