Question

A steel wire of length $20cm$ and cross-sectional area $1.0m{m}^2$ is clamped at both the ends. The coefficient of linear expansion for steel is $\alpha =$ 1.1 x 10${}^{-5}$/${}^o$C and Y$=$2 x 10${}^{11}$ N/m${}^2$. If the temperature of the wire is reduced from 40${}^o$C to 20${}^o$C, the tension developed in the wire will be

• A. $2.2x{10}^6$ N
• B. $8N$
• C. $16N$
• D. $44N$

Answer 1

Answer: (D)

$44N$

Using the relation for the change in length we get
$\Delta L=\alpha L\Delta T$
$=1.1\times {10}^{-5}\times 0.2\times (40-20)=4.4\times {10}^{-5}m$
Thus using the relation
$\Delta L=\frac{Fl}{AY}$
we get
$F=\frac{\Delta LAY}{l}=\frac{4.4\times {10}^{-4}\times 1\times {10}^{-6}\times 2\times {10}^{11}}{0.2}=44N$