Question

A bimetallic strip is formed out of two identical strips, one of copper and 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

• A. Proportional to T
• B. Inversely proportional to T
• C. Proportional to
• D. Inversely proportional to

Answer 1

Answer: (B), (D)

The correct options are
B Inversely proportional to T
D Inversely proportional to

On heating, the strip undergoes linear expansion
So after expansion length of brass strip $L_B=L_0(1+{\alpha }_B\Delta T)$ and length of copper strip $L_C=L_0(1+{\alpha }_C\Delta T)$

From the figure $L_B=(R+d)\theta \dots \dots \left(i\right)$
and $L_c=R\theta \dots \dots \left(ii\right)$ [As angle = Arc/Radius]
Dividing (i) by (ii) $\frac{R+d}{R}=\frac{L_B}{L_C}=\frac{1+{\alpha }_B\Delta T}{1+{\alpha }_C\Delta T}$
$\Rightarrow$ $1+\frac{d}{R}=(1+{\alpha }_B\Delta T)(1+{\alpha }_C\Delta T{)}^{-1}$ = $(1+{\alpha }_B\Delta T)(1-{\alpha }_C\Delta T)$ = $1+({\alpha }_B-{\alpha }_C)\Delta T$
$\Rightarrow$ $\frac{d}{R}=({\alpha }_B-{\alpha }_C)\Delta T$ or $R=\frac{d}{({\alpha }_B-{\alpha }_C)\Delta T}$ [Using Binomial theorem and neglecting higher terms]
So we can say R $\propto \frac{1}{({\alpha }_B-{\alpha }_C)}$ and R $\propto \frac{1}{\Delta T}$