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

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Answer 1 · 2018-03-14T15:00:41

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#