Question

A bimetallic strip of brass and steel, each having thickness of 1cm is kept at ${{20}^{\circ }}C$. The strip bends with the brass on the outer side. The radius of curvature of the curved surface is $({{\alpha }_{BRASS}}=18\times {{10}^{-6}}{{/}^{\circ }}C\text{, }{{\alpha }_{STEEL}}=12\times {{10}^{-6}}{{/}^{\circ }}C)$
a) 16.67m
b) 16.67cm
c) 16.67mm
d) 3.33m

Answer 1

Hint: In the above question we are basically asked to calculate the radius of curvature when there is a change in temperature. Assume that the change in temperature is ${{100}^{\circ }}C$. Therefore we will use the expression for the radius of curvature of the bimetallic strip to determine the radius.

Formula used:
$r=\dfrac{d}{({{a}_{1}}-{{a}_{2}})\Delta T}$

Complete step-by-step answer:
Let us say a bimetallic strip is made of two metals whose coefficients of thermal expansion are ${{a}_{1}}\text{ and }{{a}_{2}}$ respectively. Let us say further we subject this strip to a change in temperature $\Delta T$. This will result in a bent in the strip. Let the thickness of each metal bed. Therefore the radius of curvature for the bent metallic strip is given by,
$r=\dfrac{d}{({{a}_{1}}-{{a}_{2}})\Delta T}$
In the above question the thickness of each of the metals is given to us as 1cm. The coefficient of thermal expansion of brass as well as the steel is given to us, hence the radius of curvature from the above expression we get as,
$\begin{align}
  & r=\dfrac{d}{({{a}_{BRASS}}-{{a}_{STEEL}})\Delta T} \\
 & \Rightarrow r=\dfrac{1\times {{10}^{-2}}}{(18\times {{10}^{-6}}-12\times {{10}^{-6}})100} \\
 & \Rightarrow r=\dfrac{1\times {{10}^{-2}}}{(6\times {{10}^{-6}})100} \\
 & \Rightarrow r=\dfrac{1\times {{10}^{-2}}}{(6\times {{10}^{-4}})} \\
 & \Rightarrow r=\dfrac{1\times {{10}^{2}}}{6}=0.1666\times {{10}^{2}} \\
 & \Rightarrow r=16.67m \\
\end{align}$

So, the correct answer is “Option A”.

Note: It is to be noted that all the physical quantities have to be expressed in the SI units or else the multiplier of the answer obtained will be different. It is given that the bimetallic strip consists of brass on its outer surface. Therefore we have taken ${{\alpha }_{1}}={{\alpha }_{BRASS}}$ and ${{\alpha }_{2}}={{\alpha }_{STEEL}}$.