A ` 5 m` long cylindrical steel wire with radius ` 2xx10 ^(-3) m ` is suspended vertically from a rigid support and carries a bob of mass ` 100 kg ` at the other end . If the bob gets snapped , calculate the change in temperature of the wire ignoring radiation losses. ( Take ` g = 10 m//s ^(2) `)
( For the steel wire , young's modulus `= 2.1 xx10^(11) N//m^(2) ` , Density `= 7860 kg// m^(3) `, Specific heat `= 420 J// kg ^@ C` ) .

Answer

`4.568 xx 10^(-3) ^(@)C`

Explanation

Given

Length of the wire`l=5m`
Radius of the wire `r= 2 xx 10^(-3)m`
Density of wire `rho = 7860kg//m^(3)`
Young's modulus
`Y = 2.1 xx 10^(11) N//m^(2)` and specific heat
`S = 420J//kg-K`
Mass of wire, m (density)(volume)
=`(rho)(pir^(2)l) = (7860)pi)(2 xx 10^(-3))^(2)(5)"kg" = 0.494"kg"`
Elastic potential energy stored in the wire,
`U = (1)/(2)` (stress)(strain)`xx`(volume)
`U=(1)/(2)`
`implies U = (1)/(2)((Mg)/(pir^(2)))((Deltal)/(l))(pir^(2)l) = (1)/(2)(Mg).Deltal`
`because (Deltal = (Fl)/(AY))`
=`(1)/(2)(Mg)((Mgl))/((pir^(2))Y) = (1)/(2)(M^(2)g^(2)l)/(pir^(2)Y)`
Substituting the values, we have
`U = (1)/(2)((100)^(2)(10)^(2)(5))/((3.14)(2xx10^(-3))^(2)(2.1xx 10^(11))J = 0.9478J`
When the bob gets snapped , this energy is utilised in raising the temperature of the wire.
So, `U = "ms"Deltatheta`
`therefore Deltatheta= (U)/(ms) = (0.9478)/(0.494(420) ^(@)C "or" K`
`implies Deltatheta= 4.568 xx 10^(-3)"^(@)C`