Question

A bimetallic strip on heating has radius $R$. The coefficients of thermal expansion are ${\alpha }_1$ and ${\alpha }_2$. If the strip having greater coefficient of thermal expansion $\left({\alpha }_1\right)$ is replaced by an identical strip having coefficient of thermal expansion $3$ times its value (i.e $3{\alpha }_1$) and heated to the same temperature difference, then the new radius of the arc is
[ given, $\frac{{\alpha }_1}{{\alpha }_2}=2$ ]

• A. $\frac{R}{5}$
• B. $2R$
• C. $5R$
• D. $\frac{R}{2}$

Answer 1

The correct option is A $\frac{R}{5}$

Given:
Radius of metallic strip $=R$
Coefficient of linear expansion of strip $1$ $={\alpha }_1$
Coefficient of linear expansion of strip $2$ $={\alpha }_2$
Coefficient of linear expansion of strip $3$ with which strip $1$ is replaced, ${\alpha }_3=3{\alpha }_1$
Given: $\frac{{\alpha }_1}{{\alpha }_2}=2$
To find:
New radius of arc of bimetallic strip, $R^{{}^{\prime }}=?$
Let us suppose:
Temperature change $=\Delta T$
Thickness of each strip $=t$

We know that, radius of arc of the bimetallic strip is given by
$R=\frac{t}{({\alpha }_1-{\alpha }_2)\Delta T}$
$\Rightarrow R=\frac{t}{{\alpha }_2\Delta T}$ .......(1)
[ given, ${\alpha }_1=2{\alpha }_2$ ]
Now, after replacing strip $1$ with strip $3$,
$R^{{}^{\prime }}=\frac{t}{({\alpha }_3-{\alpha }_2)\Delta T}$
$\Rightarrow R^{{}^{\prime }}=\frac{t}{5{\alpha }_2\Delta T}$ ......(2)
[ given, ${\alpha }_3=3{\alpha }_1$ and ${\alpha }_1=2{\alpha }_2$ ]
From (1) and (2)
$R^{{}^{\prime }}=\frac{R}{5}$