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 ${\alpha }_b$. 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,
[Neglect the change in width of the rods due to expansion]

• A. Proportional to $\Delta T$
• B. Inversely proportional to $\Delta T$
• C. Proportional to $|{\alpha }_b-{\alpha }_c|$
• D. Inversely proportional to $|{\alpha }_b-{\alpha }_c|$

Answer 1

The correct option is B Inversely proportional to $\Delta T$

Let $w$ and $l_0$ be width and initial length of the two strips, respectively.
The length after heating becomes,
$l_c=l_0(1+{\alpha }_c\Delta T)$
$l_b=l_0(1+{\alpha }_b\Delta T)$
The strips bend as shown in the figure.

From the figure, the length of two strips are,
$l_c=R\theta$
$l_b=(R+w)\theta$
Dividing the above two equation
$\frac{R+w}{R}=\frac{l_b}{l_c}=\frac{1+{\alpha }_b\Delta T}{1+{\alpha }_c\Delta T}$

$=1+{\alpha }_b\Delta T\left)\right(1+{\alpha }_c\Delta T{)}^{-1}$

$=(1+{\alpha }_b\Delta T)(1-{\alpha }_c\Delta T)$ [From binomial expansion]

$=1+({\alpha }_b-{\alpha }_c)\Delta T-{\alpha }_b{\alpha }_c\Delta T^2$

${\alpha }_b{\alpha }_c\Delta T^2\to$ is a very small quantity; so we can ignore it.

So, $\frac{R+w}{R}\simeq 1+({\alpha }_b-{\alpha }_c)\Delta T$

$\Rightarrow 1+\frac{w}{R}=1+({\alpha }_b-{\alpha }_c)\Delta T$

$\Rightarrow R=\frac{w}{({\alpha }_b-{\alpha }_c)\Delta T}$

$\therefore$ Radius is inversly proportional to $\Delta T$.

Hence, $\left(B\right)$ is the correct answer.