CalorimetryHard
Question
A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6 × 10-5mm. The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (a = 12 × 10-6 /oC)
Options
A.4.0°C
B.4.5oC
C.5.0oC
D.5.5oC
Solution
6 × 10-5 = 1 × 12 × 10-6 × ᐃT
= ᐃT ⇒ ᐃT = 5oC
= ᐃT ⇒ ᐃT = 5oCCreate a free account to view solution
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