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Two rods of different metals having the ...

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.

A

`L_(2)^(1) = L_(2) [1 + alpha_(2)T - (F)/(AY_(2))]`
`L_(1)^(1) = L_(1) (1 + alpha_(1)T - (F)/(AY_(1)))`

B

`L_(2)^(1) = L_(1) [1 - alpha_(2)T + (F)/(AY_(2))]`
`L_(1)^(1) = L_(1) (1 - alpha_(1)T - (F)/(AY_(1)))`

C

`L_(2)^(1) = L_(1) [1 + alpha_(2)T + (F)/(AY_(2))]`

D

`L_(2)^(1) = L_(1) [1 - alpha_(2)T - (F)/(AY_(2))]`

Text Solution

Verified by Experts

The correct Answer is:
A
Consider the rod of length `L_(1)` Due to heating itg increases by `(Delta L_(1))_(H) = (alpha_(1) L_(1) T)` while due to compression it decreases by `(Delta L_(1))_(C) = (FL_(1)//AY_(1)),` so its final length
`L_(1) = L_(1) + (Delta L_(1))_(H) - (Delta L_(1))_(C) = L_(1) [1 + alpha_(1)T - (F//AY_(1))]`
`L_(2) = L_(2) + (Delta L_(2))_(H) - (Delta L_(2))_(C) = L_(2) [1 + alpha_(2)T - (F//AY_(2))]`
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