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 t
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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Correct Answer - C
`DeltaL= 6xx10^(-5) = Lalphatheta implies theta = (6xx10^(-5))/(1 xx 12 xx 10^(-6) =5^(@)C`
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