Question

A steel rod of cross-sectional area $4c{m}^2$ and length 2 m shrinks by 0.1 cm as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young modulus of steel = $1.9\times {10}^{11}N{m}^{-2}$.

Answer 1

A = $4c{m}^2=4\times {10}^{-4}{m}^2,L=2m$

$\Delta L=0.1cm={10}^{-3}m,Y=1.9\times {10}^{11}$

Y $=\frac{F}{A}\times \frac{L}{\Delta L}$

$\Rightarrow F=\frac{YA\Delta L}{L}$

= $\frac{1.9\times {10}^{11}\times {10}^{-4}\times {10}^{-3}}{2}$

= $3.8\times {10}^{4}N$

The tension developed is $3.8\times {10}^{4}N$.