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A steel scale is to be prepared such that the millimeter intervals are to be accurate within `6 xx10^(-5)mm`. The maximum temperature variation form the temperature of calibration during the reading of the millimeter marks is `(alpha = 12 xx 10^(-6)//"^(@)C)`A. `4.0^(@)C`B. `4.5^(@)C`C. `5.0^(@)C`D. `5.5^(@)C`. |
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Answer» Correct Answer - C `DeltaL= 6xx10^(-5) = Lalphatheta implies theta = (6xx10^(-5))/(1 xx 12 xx 10^(-6) =5^(@)C` |
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