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A steel meter scale is to be ruled so that millimeter intervals are accurate within about 5×10^–5 mm at a certain temperature. The maximum temperature variation allowable during the ruling is coefficient of linear expansion of steel = 10 ×10^–6k^–1]

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2⁰C

5⁰C

7⁰C

10⁰C

answer is B.

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Detailed Solution

$α = \frac{Δ ℓ}{ℓ Δt} \Rightarrow Δt = \frac{Δ ℓ}{ℓ α} = \frac{5 \times 10^{- 5} \times 10^{- 3}}{1 \times 10^{- 3} \times 10^{- 5}} = 5^{0} C$

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