Question

A bimetallic strip of brass and steel, each having thickness of $1cm$, is kept at ${20}^{o}C$. The strip bends with brass on outer side. The radius of curvature of curved surface is : $({\alpha }_{brass}=18\times {10}^{-6}{/}^{o}C,{\alpha }_{steel}=12\times {10}^{-6}{/}^{o}C)$

• A. 16.67 m
• B. 16.67cm
• C. 16.67 mm
• D. 3.33m

Answer 1

The correct option is A 16.67 m

radius = $r=\frac{d}{({\alpha }_1-{\alpha }_2)△t}$

where $d$ = thickness = 1 cm=$1\times {10}^{-2}m$

${\alpha }_1={\alpha }_{brass}=18\times {10}^{-6}{/}^{o}C$

${\alpha }_2={\alpha }_{steel}=12\times {10}^{-6}{/}^{o}C$

Assuming change in temperature = $△t={100}^{o}C$

$r=\frac{1\times {10}^{-2}}{(6\times {10}^{-6})100}r=16.67m$