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

Explanation:

Step 1:

According to linear expansion equation  

$L=L_{0}(1+\alpha t)$

Where,  L= length at final temperature

$L_{0}=$ length at initial temperature

t= temperature difference (final temperature – initial temperature)

α = coefficient of linear expansion

$L_{0}=1 \mathrm{cm}, \mathrm{t}=10^{\circ} \mathrm{C}, \alpha=1.1 \times 10^{-50} \mathrm{C}^{-1}$

Step 2:

By substituting the values of all variables in above equation we get

$L=1\left(1+1.1 \times 10^{-5} \times 10\right)$

$L=1+0.00011$

$L=1.00011$

Step 3:

Change in length = final length – initial length

$\Delta L=1.00011-1$

$\Delta L=0.00011 \mathrm{cm}$

Distance between $51 \mathrm{cm}$ and $50 \mathrm{cm}=1-\Delta L$

$=1-0.00011$

$=0.99989 \mathrm{cm}$