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

4.57

Explanation

Given , Length of the wire, `1=5m`
Radius of the wire , `r=2xx10^(-3)m`
Density of wire , `rho=7860kg//m^(3)`
Young's modulus , `Y=2.1xx10^(11)N//m^(2)`
And specific heat , `s=420J//kg-K`
mass of wire , m=(density)(volume)
`(rho)(pir^(2)l)`
`=(7860)(pi)(2xx10^(-3))^(2)(5)kg=0.94kg`

Elastic potential energy stored in the wire, `U=(1)/(2)` (stress) (strain) (volume)
`[because("Energy")/("Volume")=(1)/(2)xx"stress"xx"strain"]orU=(1)/(2)((Mg)/(pir^(2)))((Deltal)/(l))(pir^(2)l)`
`=(1)/(2)(Mg).Deltal(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)(2.1xx10^(11)))J=0.9478J`
When the bob gest snapped, this energy is utilized in raising the temperature of the wire.
So , `U=msDeltatheta`
`thereforeDeltatheta=(U)/(ms)=(0.9478)/(0.494(420)).^(@)CorK,Deltatheta=4.568xx10^(3).^(@)C`