At room temperature (`25^@C`) the length of a steel rod is measured using a brass centimetre scale. The measured length is 20 cm. If the scale is calibrated 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 temperature the division of scale will be more than 1 cm due to thermal expansion. Thus at higher temperatures 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 than the acktual length to be measured. For example in this case the length of each division on brass scale at `25^(@)C` is
`l_(1"div")=(1cm)[1+alpha_("br")(25-0)]`
`=1+alpha_(br)(25)`
It is given that at `25^(@)C` the length of steel rod measured is 20 cm. Actually it is not 20 cm, it is 20 divisions onthe brass scale. Now we can fine the actual length of the steel rod ast `25^(@)C` as
`l_(25.^(@)C)=(20cm)xxl_(1"div")`
or `l_("active")=20[1+alpha_("br")(25)]` .1.34
The above the expression is a general relation using which you can find the actual lengts of the objects of which lengths are measured by a metallic scale at some temperature other than the graduation temperature of the scale.