Question

Two straight metallic strips each of thickness $t$ and length $L$ are riveted together. Their coefficients of linear expansion are ${\alpha }_1$ and ${\alpha }_2$. If they are heated through temperature $\Delta \theta$, the bimetallic strip will bend to form an arc of radius

• A. $\frac{t}{({\alpha }_1+{\alpha }_2)\Delta \theta }$
• B. $\frac{t}{({\alpha }_2-{\alpha }_1)\Delta \theta }$
• C. $\frac{t}{2({\alpha }_1+{\alpha }_2)\Delta \theta }$
• D. $\frac{t}{2({\alpha }_2-{\alpha }_1)\Delta \theta }$

Answer 1

The correct option is B $\frac{t}{({\alpha }_2-{\alpha }_1)\Delta \theta }$

Given:
Thickness of metallic strips $=t$
Length of metallic strips $=L$
Coefficient of linear expansion of strip $1$ $={\alpha }_1$
Coefficient of linear expansion of strip $2$ $={\alpha }_2$
Temperature change $=\Delta \theta$
To find:
Radius of arc of bimetallic strip $=r$

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 strip $1$,
$\theta =\frac{L(1+{\alpha }_1\Delta \theta )}{r}$.......(1)
[ from formula of linear expansion ]
Similarly, for strip $2$,
$\theta =\frac{L(1+{\alpha }_2\Delta \theta )}{r+t}$.......(2)
Angle subtended by both the strips at the centre will be equal, so from (1) and (2),
$\frac{L(1+{\alpha }_1\Delta \theta )}{r}=\frac{L(1+{\alpha }_2\Delta \theta )}{r+t}$
$\Rightarrow r=\frac{t(1+{\alpha }_1\Delta \theta )}{({\alpha }_2-{\alpha }_1)\Delta \theta }$
As $1>>{\alpha }_1\Delta \theta$ [because $\alpha <<1$]
$\Rightarrow r=\frac{t}{({\alpha }_2-{\alpha }_1)\Delta \theta }$