Question

A bimetallic strip is formed out of two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are ${\alpha }_C$ and ${\alpha }_B({\alpha }_B>{\alpha }_C)$. On heating the bimetallic strip through temperature $△T$, the strip bends into an arc of a circle. Find the radius of curvature of the strip

• A. $R=\frac{2d}{({\alpha }_B-{\alpha }_C)△T}$
• B. $R=\frac{3d}{({\alpha }_B-{\alpha }_C)△T}$
• C. $R=\frac{d}{({\alpha }_B-{\alpha }_C)△T}$
• D. $R=\frac{d}{2({\alpha }_B-{\alpha }_C)△T}$

Answer 1

The correct option is C $R=\frac{d}{({\alpha }_B-{\alpha }_C)△T}$

A bimetallic strip

Both strips are identical, so length $l_o$ and width $d$ is equal.

For Copper strip,

$l_o(1+{\alpha }_C△\theta )=(R-\frac{d}{2})\theta$

For Brass strip,

$l_o(1+{\alpha }_B△\theta )=(R+\frac{d}{2})\theta$

$\Rightarrow \frac{1+{\alpha }_B△\theta }{1+{\alpha }_C△\theta }=\frac{(R+d/2)}{(R-d/2)}$

Using Componendo-Dividendo

$\Rightarrow R=\frac{d}{({\alpha }_B-{\alpha }_C)△\theta }$

Now, $△\theta =△T$

So, $R=\frac{d}{({\alpha }_B-{\alpha }_C)△T}$