Steel wire oflength 'L'at `40^(@)`C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from `40^(@)`C and `30^(@)`C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is `10^(-5)//""^(@)C`, Young's modulus of stoel is `10^(11) N//m^(2)` and radius of the wire is 1 mm. Assum that `L gt gt ` diameter of the wire. Then the value of min kg is nearly

Answer

3

Explanation

Change in length `DeltaL = L alpha DeltaT `
Also `Y = (mgL)/(A DeltaL) implies DeltaL = (mgL)/(YA) `
Equation (i) and (ii) we get `( because A = pi r^(2))`
`= (alpha Delta TY xx pi r^(2))/(g) = (10^(-5)xx (10) xx (10^(11)) xx 3.14 xx (1xx 10^(-3))^(2))/(9.8) `= 3.2 kg = 3 kg
but `dy = (dr)/(C ) . ` Therefore , `Y = (Fdr)/(pi r^(2)C dl) "or " dl = (Fdr)/(pi C Y r^(2))`
Integrating from `r= r_(1)` to `r = r_(2)` we obtain the total extension l produced in the wire by a force F which is given by
`l = int d l = (F)/(pi CY ) int_(r_(1))^(r_(2)) (dr)/(r_(2)) = (-F)/(pi CY) |(1)/(r)|_(r_(1))^(r_(2))=(F)/(pi CY) ((1)/(r_(1))-(1)/(r_(2)))= (F(r_(2)-r_(1)))/(pi CYr_(1)r_(2))`
Now F = Mg and `C = (r_(2) -r_(1))/(L) .` Hence `l = (MgL)/(pi Y r_(1)r_(2))`
Substituing the given values of `M ,L Y, r_(1)` and `r_(2)` and g = 9.8 `ms^(-2)` we have
`l =(3.14xx9.8xx10.0)/(3.14xx2xx10^(11)xx9.8xx10^(-4)xx5.0xx10^(-4))=10^(-3)m =1 ` mm