Question

The variation of lengths of two metal rods A and B with change in temperature are shown in figure. The coefficients of linear expansion ${\alpha }_A$ for the metal A and the temperature T will be :

• A. ${\alpha }_A=3\times {10}^{-6}{/}^{\circ }C,{500}^{\circ }C$
• B. ${\alpha }_A=3\times {10}^{-6}{/}^{\circ }C,{222.22}^{\circ }C$
• C. ${\alpha }_A=27\times {10}^{-6}{/}^{\circ }C,{500}^{\circ }C$
• D. ${\alpha }_A=27\times {10}^{-6}{/}^{\circ }C,{222.22}^{\circ }C$

Answer 1

The correct option is D ${\alpha }_A=27\times {10}^{-6}{/}^{\circ }C,{222.22}^{\circ }C$

$\Delta L_A=L_A{\alpha }_A\Delta T$
$\Delta L_B=L_B{\alpha }_B\Delta T$
Taking ratio of both,
$\frac{\Delta L_A}{\Delta L_B}=\frac{L_A}{L_B}\frac{{\alpha }_A}{{\alpha }_B}$
$\Rightarrow \frac{1006-1000}{1004-1002}=\frac{1000}{1002}\times \frac{{\alpha }_A}{{\alpha }_B}$
$\Rightarrow 3=\frac{1000}{1002}\times \frac{{\alpha }_A}{{\alpha }_B}$
$\Rightarrow {\alpha }_A=3\times {\alpha }_B\times \frac{1002}{1000}=3.006\times {\alpha }_B=3.006\times 9\times {10}^{-6}=27.054\times {10}^{-6}{/}^{0}C$
$T=\frac{\Delta L_A}{{\alpha }_A{L}_A}=\frac{6}{27.054\times {10}^{-6}\times 1000}={221.8}^{0}C$