Question

A bimetallic strip is formed act 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 $\alpha_{\beta}$. On heating, the temperature of the strip goes up by $\Delta T$ and the strip bends to form an arc of radius of curvature $R$. Then $R$ is:
(1) inversely proportional to $\Delta T$
(2) proportional to $\left|\alpha_{\beta}-\alpha_{c}\right|$
(3) proportional to $\Delta T$
(4) inversely proportional to $\left|\alpha_{\beta}-\alpha_{C}\right|$

• A. 1 and 2 are correct
• B. 2 and 3 are correct
• C. 1 and 4 and correct
• D. 1, 2 and 4 are correct

Answer 1

Answer: (C)

Let to be the initial length of each strip before heating.

Length after heating will be
$l_{B}=l_{0}\left(1+\alpha_{\beta} \Delta T\right)=(R+d) \theta$
and $R_{c}=l_{0}\left(1+\alpha_{c} \Delta T\right)=R \theta$
$\therefore \frac{R+d}{R}=\frac{1+\alpha_{\beta} \Delta T}{1+\alpha_{c} \Delta T} $
$\therefore 1+\frac{d}{R}=1+\left(\alpha_{\beta}-\alpha_{c}\right) \Delta T$
[from binomial expansion]
$\therefore R=\frac{d}{\left(\alpha_{\beta}-\alpha_{c}\right) \Delta T}$
Or $R \propto \frac{1}{\Delta T}$
and $\propto \frac{1}{\left|\alpha_{\beta}-\alpha_{c}\right|}$