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 $△t$, the bimetallic strip will bend to from an arc of radius

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

Answer 1

The correct option is C $\frac{t\left(2+\left({\alpha }_1+{\alpha }_2\right)\Delta t\right)}{2\left({\alpha }_2-{\alpha }_1\right)\Delta t}$

Assume:${\alpha }_2>{\alpha }_1$

So, outer more expansion

$\begin{array}{c}\theta =\frac{S}{R}=\frac{l\left(1+{\alpha }_2\Delta t\right)}{R+\frac{t}{2}}\\ =\frac{l\left(1+{\alpha }_1\Delta t\right)}{R-\frac{t}{2}}\\ \left(R+\frac{t}{2}\right)\left(1+{\alpha }_1\Delta t\right)=\left(R-\frac{t}{2}\right)\left(1+{\alpha }_2\Delta t\right)\\ \Delta R\left({\alpha }_1-{\alpha }_2\right)t=-t-\frac{t}{2}\left({\alpha }_1+{\alpha }_2\right)\Delta t\\ \Delta R\left({\alpha }_1-{\alpha }_2\right)t=-t-2\left({\alpha }_1+{\alpha }_2\right)\Delta t\\ Hence,\\ R=\frac{t\left(2+\left({\alpha }_1+{\alpha }_2\right)\Delta t\right)}{2\left({\alpha }_2-{\alpha }_1\right)\Delta t}\\ \therefore optionCiscorrectanswer.\end{array}$