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What length of brass and iron at `0^(0)C` must be used if the difference between their lengths is always `0.2m`? The values of `alpha` for brass and iron and `18 xx 10^(-6)//0 C` and `12 xx 10^(-6)//0 C` respectively,

Answer» `l_(1)alpha_(1) = l_(2)alpha_(2)` and `l_(2) - l_(1) = x , l_(1) = (l_(2)alpha_(2))/(alpha_(1))`
`l_(2) - l_(1) = x = l_(2) - (l_(2)alpha_(2))/(alpha_(1)) = l_(2)((alpha_(1) - alpha_(2))/(alpha_(1)))`
`l_(2) = (xalpha_(1))/(alpha_(1) - alpha_(2)) = (0.2 xx 12 xx 10^(+6))/((18 - 12) xx 10^(-6)) = 0.40m`
`l_(1) = (xalpha_(2))/(alpha_(1) - alpha_(2)) = (0.2 xx 18 xx 10^(-6))/((18 - 12) xx 10^(-6)) = 0.60m`


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