We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If `alpha_("iron")=1.2xx10^(-5)//K` and `alpha_("brass")=1.8xx10^(-5)//K` what should we take as length of each strip ?

Explanation

According to question, `l_("iron")-l_("brass")=10cm` . . . (1)
Constant at all temperature.
Suppose length at `0^(@)C` is `l_(0)` and it becomes l due to increase in temperature `DeltaT`
`:.l_("iron")=l_("0iron")(1+alpha_("iron")DeltaT)` and `l_("brass")=l_("0brass")(1+alpha_("brass")DeltaT)`
`:.` From equation (1),
`l_("0iron")(1+alpha_("iron")DeltaT)-l_("0brass")(1+alpha_("brass")DeltaT)=10cm`
But the difference of their lengths remains constant,
`:. L_("0iron")alpha_("iron")=l_("0brass")alpha_("brass")`
`:.(l_("0iron"))/(l_("0brass"))=(alpha_("brass"))/(alpha_("iron"))=(1.8xx10^(-5))/(1.2xx10^(-5))=3/2`
`:.l_("0iron")=3/2l_("0brass")`
Now, from equation (1),
`3/2l_("0brass")-l_("0brass")=10`
`:.1/2l_("0brass")=10`
`:.l_("0 brass")=20cm`
`:.` Original length of brass scale is 10 cm.
Now, again from equation (1),
`l_("0iron")-l_("0 brass")=10`
`l_("0iron")-20=10`
`:.l_("0iron")=30cm`
`:.` Original length of iron scale is 30 cm.