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)

Options

A. 20 cm
B. 15 cm
C. 5 cm
D. 10 cm

Answer

D

Explanation

`DeltaL_(S)-DeltaL_(B)=DeltaL`
`L_(S)=L,L_(B)=(L-5)am,alpha_(s)=12xx10^(-6)//""^(@)C`
`alpha_(B)=18xx10^(-6)//""^(@)C`
As, `alpha_(s)=(DeltaL_(s))/(L_(3)xxt_(1))rArr12xx10^(-6)=(DeltaL_(s))/(L_(3)xxt_(1))" "...(i)`
`alpha_(B)=(DeltaL_(B))/(L_(B)xxt_(2))rArr18xx10^(-6)=(DeltaL)/((L-5)xxt)" "...(ii)`
Dividing Eqs. (i) and (ii), we get
`2/3=(L-5)/(L)or L=15rArrL_(B)=15-5=10cm`