An iron metre rod is allowed an error of 1 part per million. With α = 1 0 − 5 ∘ C − 1 , the maximum temperature variation allowed is:

Let x be the magnitude of the temperature deviation. Then L Δ L max = αx ≤ 1 0 − 6 , so x ≤ 1 0 − 5 1 0 − 6 = 0.10 K .

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An iron metre rod is allowed an error of 1 part per million. With $α = 1 0^{- 5}^{\circ} C^{- 1}$ , the maximum temperature variation allowed is:

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Answer:

Solution:

Let x be the magnitude of the temperature deviation. Then $\frac{Δ L _{max}}{L} = αx \leq 1 0^{- 6}$ , so $x \leq \frac{1 0 ^{- 6}}{1 0 ^{- 5}} = 0.10 K$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

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