MHT CET · Physics · Thermal Properties of Matter
Two rods of different metals have coefficients of linear expansion ' \(\alpha_1\) ' and ' \(\alpha_2\) ' respectively. Their respective lengths are ' \(L_1\) ' and ' \(L_2\) '. At all temperatures \(\left(L_2-L_1\right)\) is same. The correct relation is
- A \(\mathrm{L}_1 \alpha_1^2=\mathrm{L}_2 \alpha_2^2\)
- B \(\mathrm{~L}_1^2 \alpha_1^2=\mathrm{L}_2^2 \alpha_2^2\)
- C \(\mathrm{~L}_1 \alpha_2=\mathrm{L}_2 \alpha_1\)
- D \(\mathrm{~L}_1 \alpha_1=\mathrm{L}_2 \alpha_2\)
Answer & Solution
Correct Answer
(D) \(\mathrm{~L}_1 \alpha_1=\mathrm{L}_2 \alpha_2\)
Step-by-step Solution
Detailed explanation
For same change in temperature the change in length is same for both the rods.
Change in length \(\Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T}\)
\(\therefore \mathrm{L}_1 \alpha_1 \Delta \mathrm{T}=\mathrm{L}_2 \alpha_2 \Delta \mathrm{T}\)
Or \(\mathrm{L}_1 \alpha_1=\mathrm{L}_2 \alpha_2\)
Change in length \(\Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T}\)
\(\therefore \mathrm{L}_1 \alpha_1 \Delta \mathrm{T}=\mathrm{L}_2 \alpha_2 \Delta \mathrm{T}\)
Or \(\mathrm{L}_1 \alpha_1=\mathrm{L}_2 \alpha_2\)
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