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

1 Answer
Mar 14, 2018

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 =(delta L)/(Lalpha)=(0.0005/1000)(1/(1.322*10^-5))=0.07^@C