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NEET 2016
Exam
NEET
Year
2016
Subject
Physics
Chapter
Thermal Properties of Matter

NEET 2016 · Physics · Thermal Properties of Matter

Coefficient of linear expansion of brass and steel rods are α _ 1...

Coefficient of linear expansion of brass and steel rods are α1{\alpha}_{1} and α2{\alpha}_{2}. Lengths of brass and steel rods are l1andl2{l}_{1} and {l}_{2}, respectively. If (l2−l1)\left({l}_{2} - {l}_{1}\right) is maintained the same at all temperatures, which one of the following relations holds good?

Options

  1. Aα1l2=α22l1{\alpha}_{1}{l}_{2} = {\alpha}_{2}^{2}{l}_{1}
  2. Bα1l22=α2l12{\alpha}_{1}{l}_{2}^{2} = {\alpha}_{2}{l}_{1}^{2}
  3. Cα12l2=α22l1{\alpha}_{1}^{2}{l}_{2} = {\alpha}_{2}^{2}{l}_{1}
  4. Dα1l1=α2l2{\alpha}_{1}{l}_{1} = {\alpha}_{2}{l}_{2}

Correct answer

D. α1l1=α2l2{\alpha}_{1}{l}_{1} = {\alpha}_{2}{l}_{2}

Step-by-step solution

The difference in length is the same, so the increase in length is equal,

∆l1=∆l2∆ {l}_{1} = ∆ {l}_{2}

l1α1∆T=l2α2∆T{l}_{1}{\alpha}_{1} ∆ T = {l}_{2}{\alpha}_{2} ∆ T

⇒l1α1=l2α2⇒ {l}_{1}{\alpha}_{1} = {l}_{2}{\alpha}_{2}.

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