Two rods of different metals having the same area of cross section A are placed between the two massive walls as shown is Fig. The first rod has a length `l_1`, coefficient of linear expansion `alpha_1` and Young's modulus `Y_1`. The correcsponding quantities for second rod are `l_2,alpha_2` and `Y_2`. The temperature of both the rods is now raised by `t^@C`.
i. Find the force with which the rods act on each other (at higher temperature) in terms of given quantities.
ii. Also find the length of the rods at higher temperature.

Explanation

Let `t^@C=`increase in the temperature.
Increase in length of first rod`=l_1alpha_1t`
Increase in length of second rod`=l_2alpha_2t`
Total extension in length due to rise in temperature
`=l_1alpha_1t+l_2alpha_2t=(l_1alpha_1+l_2alpha_2)t` ..(i)
since the walls are rigid, this increase in length will not happen. This will be compensated by equal and opposite forces F, F producing decrease in the lengths of the rods due to elasticity.
Decrease in length of first rod`=(Fxxl_1)/(Y_1xxA)`
And decrease in length of second rod `=(Fxxl_2)/(Y_2xxA)`
Total decrease in length due to elastic force
`=(F)/(A)((l_1)/(Y_1)+(l_2)/(Y_2))` ..(ii)
From Eqs. (i) and (ii) we have
`(F)/(A)((l_1)/(Y_1)+(l_2)/(Y_2))=(l_1alpha_1+l_2alpha_2)t`
or `F=(A(l_1alpha_1+l_2alpha_2)t)/((l_1)/(Y_1)+(l_2)/(Y_2))` ..(iii)
ii. Length of the first rod`=`original length`+`increase in length due to temperature `-` decrease in length due to force
`=(l_1+l_1alpha_1t-(F)/(A)(l_1)/(Y_1))`
and length of second rod`=l_2+l_2alpha_2t-(F)/(A)(l_2)/(Y_2)`
The total length is same `=l_1+l_2` at all temperatures.