Q. A steel wire of length 20 cm and uniform cross-section 1 $mm^{2}$ is tied rigidly at both the ends. The temperature of the wire is altered from $40℃\ to\ 20℃.$ Coefficient of linear expansion of steel is $\alpha = 1.1\ \times 10^{- 5}℃^{- 1}$ and Y for steel is $2.0\ \times 10^{11}\ Nm^{- 2};$ the tension in the wire is
A) $2.2\ \times 10^{6}\ N$
B) 16 N
C) 8 N
D) 44 N
Explanation
(d) Increase in length due to rise in temperature $\mathrm{\Delta}L = aL\mathrm{\Delta}T$
As $Y = \frac{FL}{A\mathrm{\Delta}L},\ so,\ \ F = \frac{YA\mathrm{\Delta}l}{L} = \ \frac{YA \times aL\mathrm{\Delta}T}{L} = YAa\mathrm{\Delta}T$
$\therefore\ \ \ \ \ \ F = 2 \times 10^{11} \times 10^{- 6} \times 1.1 \times 10^{- 5} \times 20 = 44\ $N.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.