A body of mass `3.14 kg` is suspended from one end of a wire of length `10 m`. The radius of cross-section of the wire is changing uniformly from `5xx10^(-4) m` at the top (i.e. point of suspension) to `9.8xx10^(-4) m` at the bottom . Young's modulus of elasticity is `2xx10^(11) N//m^(2)`. The change in length of the wire is

Options

A. `4xx10^(-3) m`
B. `3xx10^(-3) m`
C. `10^(-3) m `
D. `2xx10^(-3) m `

Answer

C

Explanation

Change in length of element is
Cross-sectional area of the element is
`A = pi[a+(b-a)/Lx]^(2)`
Where , ` a =( b-a)/(L)x` is radius of the wire at the location of element .
`d(Deltal) = F/(pi[a+(b-a)/Lx]^(2)Y )dx`
Total change in length of wire is
`Deltal = F/(piY) int_(o)^(L) 1/(a+(b-a)/Lx)^(2)dx`
Let `a+(b-a)/Lx =t`
`rArr (b-a)/Ldx =dt rArr dx = (L)/(b-a)dt`
When `x=o` , `t =a` and when `x= L , t = b`
`:. Deltal = (F)/(piY) int_(a)^(b) t^(-2).(L )/(b-a)dt`
`= (FL)/(piY(b-a))(-1)(1/b-1/a)`
` = (FL)/(piY(b-a)).(b-a)/(ab)= (FL)/piYab`
`= ((3.14)(9.8)(10))/((3.14)(2xx10^(11))(5xx10^(-4))(9.8xx10^(-4))`
`= 10^(-3) m rArr :. DeltaL = 10^(-3) m`