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 ∆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 $|\alpha_{B} - \alpha_{C}|$
• D. Inversely proportional to $|\alpha_{B} - \alpha_{C}|$

Answer 1

Answer: (C)

Proportional to $|\alpha_{B} - \alpha_{C}|$

Answer 2

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 $L_{B} = (R + d)\theta$ ......(i)

and $L_{c} = R\theta$ ......(ii) [As angle = Arc/Radius]

Dividing (i) by (ii) $\frac{R + d}{R} = \frac{L_{B}}{L_{C}} = \frac{1 + \alpha_{B}\Delta T}{1 + \alpha_{C}\Delta T}$

⇒$1 + \frac{d}{R} = (1 + \alpha_{B}\Delta T)(1 + \alpha_{C}\Delta T)^{- 1}$

= $(1 + \alpha_{B}\Delta T)(1 - \alpha_{C}\Delta T)$ = $1 + (\alpha_{B} - \alpha_{C})\Delta T$

⇒ $\frac{d}{R} = (\alpha_{B} - \alpha_{C})\Delta T$ 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 $\propto \frac{1}{\Delta T}$