A steel wire of uniform area 2 m$m^{2}$ is heated up to 50$℃$ and is stretched by tying its ends rigidly. The change in tension when the temperature falls from 50$℃$ to 30$℃$ is (Take $Y = 2 imes 10^{11}Nm^{- 2},\ \alpha = 1.1\ imes 10^{- 5}℃^{- 1})$

Question

A steel wire of uniform area 2 m$m^{2}$ is heated up to 50$℃$ and is stretched by tying its ends rigidly. The change in tension when the temperature falls from 50$℃$ to 30$℃$ is (Take $Y = 2 	imes 10^{11}Nm^{- 2},\ \alpha = 1.1\ 	imes 10^{- 5}℃^{- 1})$

Dr. Abubacker · Accepted Answer

(c) $\mathrm{\Delta}L = \ \alpha L(\mathrm{\Delta}T) = \frac{F}{A}\frac{L}{Y}$
$	herefore\ F = \ \alpha(\mathrm{\Delta}T)AY$
= 1.1$	imes 10^{- 5} 	imes (50 - 30) 	imes 2 	imes 10^{- 6} 	imes 2 	imes 10^{11}$
=88 N

Q. A steel wire of uniform area 2 m$m^{2}$ is heated up to 50$℃$ and is stretched by tying its ends rigidly. The change in tension when the temperature falls from 50$℃$ to 30$℃$ is (Take $Y = 2 \times 10^{11}Nm^{- 2},\ \alpha = 1.1\ \times 10^{- 5}℃^{- 1})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?