A metal meter scale gives correct measurement at  ${40}^{0}C$  it is generally used at a temperature of $0^{0}C$ . Find the correction to be made for every meter  $\left(\alpha ={10}^{-6}{/}^{0}C\right)$

$$\begin{gathered} change in length is \Delta l=l\alpha \Delta t \\ l is initial length of rod =1m;\alpha is coefficient of linear expansion={10}^{-6}/{}^{0}C \\ \Delta t is change in temperatue; decreased from {40}^{0}C to 0^{0}C \\ \Delta l=1\times {10}^{-6}\times 40 \\ \Delta l=4\times {10}^{-5}m \end{gathered}$$

If $t_2<t_1$  so correction must be deducted.