Question

A bimetallic strip is formed out of two identical length strips of which one is copper and the other is brass. The coefficient 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 which of the following is/are correct ?

• A. $R\propto \Delta T$
• B. $R\propto \frac{1}{\Delta T}$
• C. $R\propto |{\alpha }_B-{\alpha }_C|$
• D. $R\propto \frac{1}{|{\alpha }_B-{\alpha }_C|}$

Answer 1

Answer: (B), (D)

$R\propto \frac{1}{|{\alpha }_B-{\alpha }_C|}$

Given:
Coefficient of linear expansion of copper $={\alpha }_C$
Coefficient of linear expansion of brass $={\alpha }_B$
Temperature change $=\Delta T$
To find:
Radius of arc of bimetallic strip $=R$
Let us assume:
Thickness of strips $=t$
Length of strips $=L$

Before heating -


After heating -


Let the angle subtended by the arc formed at the centre be $\theta$.
We know,
$\theta =\frac{l}{r}$, where $l$ is the length of arc and $r$ is radius of arc.
For copper,
$\theta =\frac{L(1+{\alpha }_C\Delta T)}{R}$.......(1)
[ from formula of linear expansion ]
Similarly, for brass,
$\theta =\frac{L(1+{\alpha }_B\Delta T)}{R+t}$.......(2)
Angle subtended by both the strips at the centre will be equal, so from(1) and (2),
$\frac{L(1+{\alpha }_C\Delta T)}{R}=\frac{L(1+{\alpha }_B\Delta T)}{R+t}$
$\Rightarrow R=\frac{t(1+{\alpha }_C\Delta T)}{(|{\alpha }_B-{\alpha }_C|)\Delta T}$
As $1>>{\alpha }_C\Delta T$ [$\alpha$ is very small]
$\Rightarrow R=\frac{t}{(|{\alpha }_B-{\alpha }_C|)\Delta T}$
$\Rightarrow R\propto \frac{1}{\Delta T}$ and $R\propto \frac{1}{|{\alpha }_B-{\alpha }_C|}$