Coefficient of linear expansion of brass and steel rods are `alpha_(1) " and " alpha_(2)`. Lengths of brass and steel rods are `l_(1) " and " l_(2)` respectively. If `(l_(2)-l_(1))` is maintained same at all temperatures, which one of the following relations holds good?

Options

A. `alpha_(1)l_(2)^(2)=alpha_(2)l_(1)^(2)`
B. `alpha_(1)^(2)l_(2)=alpha_(2)^(2)l_(1)`
C. `alpha_(1)l_(1)=alpha_(2)l_(2)`
D. `alpha_(1)l_(2)=alpha_(2)l_(1)`

Answer

C

Explanation

Let, lengths of brass rod at temperatures `t_(1) " and " t_(2)` are `l_(1) " and " l_(1)^(.)` respectively.
`therefore " "l_(1)^(.)=l_(1){1+alpha_(1)(t_(2)-t_1)}`
Again, let lengths of steel rod at temperatures `t_(1) " and " t_(2)` are `l_(2) " and " l_(2)^(.)` respectively.
`therefore l_(2)^(.)=l_(2){1+alpha_(2)(t_(2)-t_(1))}`
According to the question, `l_(2)-l_(1)=l_(2)^(.)-l_(1)^(.)`
or, `l_(2)-l_(1)=l_(2){1+alpha_(2)(t_(2)-t_(1))}-l_(1){1+alpha_(1)(t_(2)-t_(1))}`
or, `l_(2)-l_(1)=l_(2)+l_(2)alpha_(2)(t_(2)-t_(1))-l_(1)-l_(1)alpha_(1)(t_(2)-t_(1))`
or, `l_(2)-l_(1)=(l_(2)-l_(1))+(t_(2)-t_(1))(l_(2)alpha_(2)-l_(1)alpha_(1))`
or, `(t_(2)-t_(1))(l_(2)alpha_(2)-l_(1)alpha_(1))=0`
as `(t_(2)-t_(1)) ne 0`
`therefore " " (l_(2)alpha_(2)-l_(1)alpha_(1))=0 " or, " l_(1)alpha_(1)=l_(2)alpha_(2)`
The option (c) is correct.