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Linear thermal expansion

Distinguish a small length change from the final length and convert metres to millimetres correctly.

Question

A rod is initially 2.0 m long and has linear expansion coefficient 1.2×10−5 K−11.2\times10^{-5}\,\mathrm{K}^{-1}. It warms by 50 K. What is its increase in length?

  1. a.0.12 mm
  2. b.1.2 mm
  3. c.12 mm
  4. d.2.0012 m

Correct answer

B. 1.2 mm

Full reasoning

  1. 1Use ΔL=αL0ΔT\Delta L=\alpha L_0\Delta T, with the initial length rather than an unknown final length.
  2. 2ΔL=(1.2×10−5)(2.0)(50)=0.0012 m\Delta L=(1.2\times10^{-5})(2.0)(50)=0.0012\,\mathrm{m}.
  3. 3Multiply metres by 1000 to get 1.2 mm.
  4. 4The fractional increase is 0.0006, which is small and consistent with the linear approximation. A temperature difference of 50 °C would have the same numerical value as 50 K.

Why each choice is right or wrong

Choice A

This is a factor of ten too small; 0.0012 m equals 1.2 mm because there are 1000 millimetres in one metre.

Choice B

Correct. Multiplying the coefficient, initial length and temperature change gives 0.0012 m, which is 1.2 mm.

Choice C

This is a factor of ten too large: 0.0012 m converts to 1.2 mm, not 12 mm.

Choice D

This is the final length, whereas the question asks for the increase.

Related formula

Linear expansion

ΔL = α L0 ΔT

Assumptions: The fractional expansion is small and α is approximately constant.

Related topic and practice

Thermodynamics Domain GuideThermodynamics and Heat EnginesCarnot Engine Work OutputHeat flow through a thicker wallCreate a free account to practice

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Sources

Original authored exercise; no database question ID and no protected or official exam-bank content. Specialist review pending.

  1. University Physics Volume 2: Linear thermal expansion — OpenStax. Accessed 2026-09-13. Textbook used as a reference for cited concepts, without reproducing textbook prose or illustrations. Exercise wording and numbers are original; no official exam questions are reproduced.
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Sources last checked 2026-09-13

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