At room temperature `(25^(0)C)` the length of a steel rod is measured using a brass centimeter scale. The measured length is 10 cm. If the scale is calibrated to read to read accurately at temperature `0^(@)C`, find the actual length of steel rod at room temperature.

Explanation

The brass scale is calibrated to read accurately at `0^(@)C`, this means at `0^(@)C`, each division of scale has exact 1 cm length. Thus at higher temmperature the division length of scale will be more than 1 cm due to thermal expansion. Thus at higher temperature the scale reading for length measurement is not appropriate and as at higher temperature the division length is more, the length this scale reads will be lesser then the actual length to be measured. For example in this case the length of each division on brass scale at `25^(0)C` is
`l_(1" div")=(1cm)[1+alpha_("br")(25-0)]=1+alpha_("br")(25)`
It is given that at `25^(0)C` the length of steel rod measured is 10 cm. Actually it is not 10 cm, it is 10 divisions on the brass scale. Now we can find the actual length of the steel rod at `25^(@)C` as
`l_(25^(0)C)= (10cm)xx l_("1 div")(or) l_("act")=10[1+alpha_("br")(25)]`
The above expression is a general relation using which you can find the actual length of the objects of which lengths are measured by a metallic scale at some temperature other than the graduation temperature of the scale.