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It is required to prepare a steel metre scale such that the millimetre intervals are to be accurate within 0.0005 mm at a certain temperature. Maximum temperature variation
allowable during the rulings of millimetre marks is :
$(α_{steel} = 13.22 \times 10^{- 6} / {}^{o}C)$

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detailed solution

Correct option is A

Here,l=1 mm,∆l=0.0005 mm As, ∆l=l α ∆T So, ∆T=∆lαl=0.000513.22×10-6×1=37.8C o.

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It is required to prepare a steel metre scale such that the millimetre intervals are to be accurate within 0.0005 mm at a certain temperature. Maximum temperature variationallowable during the rulings of millimetre marks is :(αsteel=13.22×10-6/C o)

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Ready to Test Your Skills?

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It is required to prepare a steel metre scale such that the millimetre intervals are to be accurate within 0.0005 mm at a certain temperature. Maximum temperature variation
allowable during the rulings of millimetre marks is :
$(α_{steel} = 13.22 \times 10^{- 6} / {}^{o}C)$