Question

A mass of $2kg$ is suspended from a fixed point by a wire of length $3m$ and diameter $0.5mm$. Initially the wire is just unstressed, the mass resting on a fixed support. By how much must the temperature (in ${}^{\circ }C$ ) fall if the mass is to be entirely supported by the wire ?
$(Y$ of wire $=206GPa,\alpha =11\times 10^{-6}{}^{\circ }{C}^{-1})$

Answer 1

Answer: 44.00

Here, mass $=m=2kg$
Length of wire through which mass is suspended
$=\ell =3m$
Diameter of wire $=d=5\times 10^{-4}m$
$Y_{\text{wire }}=200GPa=200\times 10^{9}N/m^2$
Coefficient of linear expansion of wire
$={\alpha }_{\text{wire }}=11\times 10^{-6}{}^{\circ }{C}^{-1}$
$Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{F/A}{\Delta \ell /\ell }$
$\Rightarrow \Delta \ell =\frac{F\ell }{AY}$
In this case, force which stretches the wire is the weight ($mg$) of body,
Let temperature fall required so that mass is entirely supported by wire be $\Delta T$
$\Delta {\ell }_{\text{wire }}=\frac{{\ell }_{\text{wire }}\times F}{A_{\text{wire }}{Y}_{\text{wire }}}$
$=\frac{3\times mg}{\pi {\left(\frac{d}{2}\right)}^2\times 206\times 10^9}$
$=\frac{3\times 2\times 9.8}{3.14\times {\left(\frac{5}{2}\times 10^{-4}\right)}^2\times 206\times 10^9}$
$\Delta {\ell }_{\text{wire }}=\frac{6\times 9.8}{3.14\times 6.25\times 206\times 10^{9-8}}$
$\Delta {\ell }_{\text{wire }}=\frac{58.8}{4642.75\times 10}$
$=1.454\times 10^{-3}m$
Also, $\Delta {\ell }_{\text{wire }}=\ell \alpha \Delta T$
$1.454\times 10^{-3}=3\times 10^{-6}\times 11\times \Delta T$
$\Delta T=\frac{1.454\times 10^{-3}}{33\times 10^{-6}}$
$=44.06^{\circ }C$