Question

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.

Answer 1

Given:
Temperature at which the steel metre scale is calibrated, t₁ = 20^oC
Temperature at which the scale is used, t₂ = 10^oC
So, the change in temperature,

\[\Delta\]t = 20^o

\[-\]10^o) C

The distance to be measured by the metre scale, L_o = 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 L_​2, 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