Question

A brass rod of length $40cm$ is joined to a copper rod of length $50cm$. The two rods are of the same thickness and at an initial temperature of ${40}^{o}C$. What is the change in length of the combined rod, when the same is heated to ${280}^{o}C.$ Coefficients of linear expansion of brass and copper are $1.9\times {10}^{-5}{}^o{C}^{-1}$ and $1.7\times {10}^{-5}{}^o{C}^{-1}$ respectively.

Answer 1

For brass rod :$\alpha =1.9\times {10}^{-5}{}^o{C}^{-1}$
$l_1=40cm;\Delta T=280-40={240}^{o}C$

The length of the brass rod at ${280}^{o}C$ is given by $l_2=l_1(1+\alpha \Delta T)$
$=40\times (1+1.9\times {10}^{-5}\times 240)$
$=40.1824cm$

For copper rod : ${\alpha }^{|}=1.7\times {10}^{-5}{}^o{C}^{-1}$
$l_1^{|}=50cm;\Delta T^{|}=280-40={240}^{o}C$

The length of the steel rod at ${280}^{o}C$ by given by :
$l_2^{|}=l_1^{|}(1+{\alpha }^{|}\Delta T)$
$=50\times (1+1.7\times {10}^{-5}\times 240)$
$=50.2040cm$

Therefore, the length of the combined rod at ${280}^{o}C,$
$l_1^{\prime }+l_2^{\prime }=40.1824+50.2040$
$=90.3864cm$
As the length of the combined rod at ${40}^{o}C$ is 50+40 i.e., 90 cm, the change in length of the combined rod at ${280}^{o}C=90.3846-90=0.3864cm$