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, t_{1} = 20^{o}C

Temperature at which the scale is used, t_{2} = 10^{o}C

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 = L_{1} ∝_{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 (L_{2}) will be,

L_{2} = 1 cm

\[-\] 0.00011 cm = 0.99989 cm