# It is required to prepare a steel meter scale, such that the mm intervals are to be accurate within 0.0005mm at a certain temperature. Determine the max. temp. variation allowable during the rulings of mm marks? Given α for steel = 1.322 x 10-5 0C-1

If the change in length is $\delta L$ of a meter scale of original length $L$ due to change in temperature $\delta T$,then,
$\delta L = L \alpha \delta T$
For, $\delta L$ to be maximum,$\delta T$ will also have to be maximum,hence,
$\delta T = \frac{\delta L}{L \alpha} = \left(\frac{0.0005}{1000}\right) \left(\frac{1}{1.322 \cdot {10}^{-} 5}\right) = {0.07}^{\circ} C$