A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10–5 °C–1.
Solution
Given:
Temperature at which the steel metre scale is calibrated, t1 = 20oC
Temperature at which the scale is used, t2 = 10oC
So, the change in temperature,
\[\Delta\]t = 20o
\[-\]10o) C
The distance to be measured by the metre scale, Lo = 51
-50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel,
\[\alpha_{steel}\]= 1.1 × 10–5 °C–1
Let the new length measured by the scale due to expansion of steel be L2, Change in length is given by,
ΔL = L1 ∝steel Δt
⇒`ΔL = 1 × 1.1 × 10^-5 × 10`
\[ \Rightarrow ∆ L = 0 . 00011 cm \]
As the temperature is decreasing, therefore length will decrease by
\[∆ L\]
Therefore, the new length measured by the scale due to expansion of steel (L2) will be,
L2 = 1 cm
\[-\] 0.00011 cm = 0.99989 cm
