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Thermal Properties of Matter · NEET Preparation Physics

A brass rod of length 50text{ cm} and diameter 3.0text{ mm} is joined to a steel rod of the same length and diameter. What is the change in length of…

Practice question from Thermal Properties of Matter (NEET Preparation Physics), with the correct answer and a step-by-step solution.

medium · MCQ · Q-B05DB7

A brass rod of length 50 cm50\text{ cm} and diameter 3.0 mm3.0\text{ mm} is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod when heated from 0∘C0^\circ\text{C} to 100∘C100^\circ\text{C}? (Coefficient of linear expansion for brass =2.0×10−5 K−1= 2.0 \times 10^{-5}\,\text{K}^{-1}, for steel =1.2×10−5 K−1= 1.2 \times 10^{-5}\,\text{K}^{-1})

  1. A.1.2 mm
  2. B.1.6 mmCorrect answer
  3. C.0.8 mm
  4. D.2.4 mm

Solution

The total change in length is the sum of the changes in length of both rods. ΔL=ΔLbrass+ΔLsteel=L1α1ΔT+L2α2ΔT=(0.5 m)(2.0×10−5+1.2×10−5)(100 K)=0.5×3.2×10−5×100=1.6×10−3 m=1.6 mm\Delta L = \Delta L_{\text{brass}} + \Delta L_{\text{steel}} = L_1 \alpha_1 \Delta T + L_2 \alpha_2 \Delta T = (0.5\text{ m})(2.0 \times 10^{-5} + 1.2 \times 10^{-5})(100\text{ K}) = 0.5 \times 3.2 \times 10^{-5} \times 100 = 1.6 \times 10^{-3}\text{ m} = 1.6\text{ mm}.

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