Question

We would like to increase the length of a $15cm$ long copper rod of cross-section $4m{m}^2$ by $1mm.$ The energy absorbed by the rod if it is heated is $E_1$ . The energy absorbed by the rod if it is stretched slowly is $E_2.$ Then if $\frac{E_1}{200E_2}=\frac{250}{x}$ find $x$ . [Various parameters of Copper are : Density $=9\times 10^{3}kg{m}^{-3},$ Thermal coefficient of linear expansion $=16\times 10^{-6}{K}^{-1}$ Young's modulus $=135\times 10^{9}Pa$ , Specific heat $=400Jk{g}^{-1}{K}^{-1}$ ]

Answer 1

Answer: 100

Temperature is increased by $\Delta \theta$ then
$\Delta l=l\alpha \Delta \theta \Rightarrow \Delta \theta =\frac{\Delta l}{l\alpha }$
$E_1=\left(\rho Al\right)S\Delta \theta =\rho AlS\frac{\Delta l}{l\alpha }$
when stretched. Stress $=Y\frac{\Delta l}{l}$
$E_2=\frac{1}{2}\left(Y\frac{\Delta l}{l}\right)\left(\frac{\Delta l}{l}\right)\times Al=\frac{Y{\left(\Delta l\right)}^{2}A}{2l}$
$\frac{E_1}{E_2}=\frac{\rho AlS\Delta l\times 2l}{\alpha l\times Y{\left(\Delta l\right)}^{2}A}=\frac{2\rho Sl}{\alpha (\Delta l)Y}=500$
Required value is, $\frac{E_1}{200E_2}$
$=\frac{E_1}{200E_2}=\frac{500}{200}=\frac{250}{100}$