The length of s steel rod exceeds that of a brass rod by 5 cm. If the difference in their lengths remains same at all temperature, then the length of brass rod will be: (`alpha` for iron and brass are `12xx10^(-6)//^(@)C` and `18xx10^(-6)//^(@)C`, respectively)

Explanation

Given, `Deltal_(St)-Deltal_(Br)=Deltal,`
`" " l_(St)=l implies l_(Br)=(l-5)`cm
(`because` steel rod is `5` cm longer than that of a brass rod)
`alpha_(St)=12xx10^(-6)//"^(@)C` and `alpha_(Br)=18xx10^(-6)//"^(@)C`
We know that, `Deltal_(St)=l_(St) alphat_(l)`
`therefore" " alpha_(St)=(Deltal_(St))/(l_(St)xxt_(l))`
`implies" " 12xx10^(-6)=(Deltal)/(lxxt)`
`"Similary," " " Deltal_(Br)=l_(Br)alphat_(2)`
`alpha_(Br)=(Deltal_(Br))/l_(Brxxt_(2))``implies" " 18xx10^(-6)=(Deltal)/((l-5)xxt)`
Deviding Eq. (i) by Eq , (ii), we get
`(12xx10^(-6))/(18xx10^(-6))=(Deltal//lxxt)/(Deltal//(l-5)xxt)implies(2)/(3)=(l-5)/(l)`
`2l=3l-15impliesl=15`
`So," " l_(Br)=l5=15-5=10` cm