Question

We would like to increase the length of a $15cm$ long copper rod of cross-section $4mm^{2}$ 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 $\frac{E_{1}}{200 E_{2}}=\frac{250}{x}$ find $x$ . [Various parameters of Copper are : Density $=9\times 10^{3}kgm^{- 3},$ Thermal coefficient of linear expansion $=16\times 10^{- 6}K^{- 1}$ Young's modulus $=135\times 10^{9}Pa$ , Specific heat $=400Jkg^{- 1}K^{- 1}$ ]

Answer 1

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