Question

Figure shows the graphs of elongation versus temperature for two different metals. If these metals be employed to form a straight bimetallic strip of thickness 6 cm and heated it bends in the form of arc,the radius of curvature changing with temperature approximately is as shown in the figure. The linear expansivities of the two metals are:

• A. $24\times {10}^{-6}{/}^{\circ }C$ and $12\times {10}^{-6}{/}^{\circ }C$
• B. $20\times {10}^{-6}{/}^{\circ }C$ and $10\times {10}^{-6}{/}^{\circ }C$
• C. $18\times {10}^{-6}{/}^{\circ }C$ and $9\times {10}^{-6}{/}^{\circ }C$
• D. $16\times {10}^{-6}{/}^{\circ }C$ and $8\times {10}^{-6}{/}^{\circ }C$

Answer 1

The correct option is D $16\times {10}^{-6}{/}^{\circ }C$ and $8\times {10}^{-6}{/}^{\circ }C$

Given : $L=6$ cm

For metal 1 : Slope $=\frac{\Delta L}{\Delta T}=\frac{4\times {10}^{-6}m}{4^{o}C}=1\times {10}^{-4}$ $cm{/}^{o}C$

$\therefore$ From $\frac{\Delta L}{\Delta T}=L\alpha$

We get $1\times {10}^{-4}cm{/}^{o}C=\left(6cm\right)\alpha$ $⟹\alpha =16\times {10}^{-6}$ ${/}^{o}C$

For metal 2 : Slope $=\frac{\Delta L}{\Delta T}=\frac{2\times {10}^{-6}m}{4^{o}C}=0.5\times {10}^{-4}$ $cm{/}^{o}C$

$\therefore$ From $\frac{\Delta L}{\Delta T}=L\alpha$

$\therefore$ $0.5\times {10}^{-4}cm{/}^{o}C=\left(6cm\right)\alpha$ $⟹\alpha =8\times {10}^{-6}$ ${/}^{o}C$