Q. A copper wire 2 m long is stretched by 1 mm. If the energy stored in the stretched wire is converted to heat, calculate the rise in temperature of the wire. (Given, $Y = 12 \times 10^{11}\ dyne\ cm^{- 2},\ $density of copper= $9\ gcm^{- 3}\ $and specific heat of copper = $0.1\ cal\ g^{- 1}℃^{- 1})$
A) $252℃$
B) (1/ 252)$℃$
C) $1000℃$
D) 2000$℃$
Explanation
(b) $E = \frac{1}{2}\ \frac{YA/\mathrm{\Delta}l^{2}}{l}$
But $m = Ald\ or\ A = \frac{m}{ld}\ $
$\therefore\ \ E = \frac{Ym\mathrm{\Delta}l^{2}}{2l^{2}d}\ $
E in calorie = $\frac{Ym\mathrm{\Delta}l^{2}}{2l^{2}d}\ $
Now, $mS\theta = \frac{Ym\mathrm{\Delta}l^{2}}{2l^{2}dJ}\ \ \ \ \ or\ \theta = \frac{Y\mathrm{\Delta}l^{2}}{2l^{2}dJS}$
Or $\theta = \ \frac{12 \times 10^{11} \times 10^{- 1} \times 10^{- 3} \times 10^{- 3}}{2 \times 2 \times 2 \times 9 \times 10^{3} \times 4.2 \times 0.1 \times 10^{3}}$
= $\frac{12 \times 10^{5}}{72 \times 42 \times 10^{5}} = \frac{1}{252}℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.