Question

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.

Accepted Answer

The brass scale is calibrated to read accurately at $0 ° C$ . This means at $20 cm$ , each division of scale has exact $0 ° C$ length. Thus at higher temperature 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 than the actual length to be measured. For illustration in this case the length of each division on brass scale at $25 ° C$ is

$l_{1 div} = (1 cm) [1 + α_{b r} (25 - 0)] = 1 + α_{b r} (25)$

where $α_{b r}$ is coefficient of linear expansion of brass scale. 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 on the brass scale. Now, we can find the actual length of the steel rod at $25 ° C$ as

$l_{25 ° C} = (20 cm) \times l_{1 div}$

or $l_{act} = 20 [1 + α_{br} (25)] . . . (i)$

The above expression is a general relation using which you can find the actual lengths of the objects of which lengths are measured by a metallic scale at some temperature other than the graduation temperature of the scale.

B M Sharma Solutions for Chapter: Thermometry, Thermal Expansion and Calorimetry, Exercise 1: ILLUSTRATION

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Thermometry, Thermal Expansion and Calorimetry, Exercise 1: ILLUSTRATION

Questions from B M Sharma Solutions for Chapter: Thermometry, Thermal Expansion and Calorimetry, Exercise 1: ILLUSTRATION with Hints & Solutions

Learn and Prepare for any exam you want !