Question

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

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