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.

Answer

`F = (AT(L_(1)alpha_(1)+L_(2)alpha_(2))Y_(1)Y_(2))/(L_(1)Y_(2)+L_(1)Y_(1))`, Length of the rod `= L_(1) +(L_(1)L_(2)T(Y_(1)alpha_(1)-Y_(2)alpha_(2)))/(L_(1)Y_(2)+L_(2)Y_(1))`, Length of second rod `= L_(2)+(L_(1)L_(2)T(Y_(2)alpha_(2)-Y_(1)alpha_(1)))/(L_(1)Y_(2)+L_(2)Y_(1))`

Explanation

`L_(A) = L_(1) +x`
`L_(B) = L_(2) - x`
`Y = (F xx l)/(Ae)`

`e = (F xxl)/(YA) =` extension or compression due to force
`e = L_(1) alpha_(1) T -x =(FL_(1))/(Y_(1)A) ……….(1)`
`e = L_(2) alpha_(2) T +x = (FL_(2))/(Y_(2)A) ....(2)`
By adding equation (1) and equation (2)
We get `F = (AT(L_(1)alpha_(1)+L_(2)alpha_(2))Y_(1)Y_(2))/(L_(1)Y_(2)+L_(2)Y_(1))`
By dividing (1) and (2)
We get `x = (L_(1)L_(2)T(Y_(1)alpha_(1)-Y_(2)alpha_(2)))/(L_(1)Y_(2)+L_(2)Y_(1))`
So, `L_(A) = L_(1)+x, L_(B) = L_(2) - x`