A thin rod of negligible mass and area of corss-section `2xx10^(-6)m^(2)`, suspended vertically from one end, has a length of `0.5m` at `200^(@)C`. The rod is cooled to `0^(@)C`, but length is kept same by attaching a mass at the lower end. The value of this mass is: (Yound's modulus=`10^(11)N//m^(2)`, coefficient of linear expansion = `10^(-5)K^(-1)`).

Answer

(i) 40 (ii) 0.1

Explanation

The change in length due to decrease in temperature,
`Delta1_(l)=LalphaDeltatheta=(0.5)(10^(-5))(0-100)`
`Delta1_(l)=-0.5xx10^(-3)m` ... (i)
Negative sign implies that length is decreasing. Now, let M be the mass attached to the lower end. Then, change in length due to suspension of load is
`Deltal_(2)=((Mg)L)/(AY)=((M)(10)(0.5))/((4xx10^(-6))(10^(11)))["Actully"L=0.5-(0.5xx10^(-3))~~0.5m`]
`Deltal_(2)=(1.25xx10^(-5))M`... (ii)
`Deltal_(1)+Deltal_(2)=0or(1.25xx10^(-5))M=(0.5xx10^(-3))thereforeM=((0.5xx10^(-3))/(1.25xx10^(-5)))kgorM=40kg`
(ii) Energy stored,
At 0°C the natural length of the wire is less than its actual length, but since a mass is attached at its lower end, an elastic potential energy is stored in it. This is given by
`U=(1)/(2)((AY)/(L))(Deltal)^(2)` ... (iii)
Here , `Deltal=|Deltal_(1)Deltal_(2)=0.5xx10^(-3)m`
Substituting the value , `U=(1)/(2)((4xx10^(-6)xx10^(11))/(0.5))(0.5xx10^(-3))^(2)=0.1J`