Question

A mass of $2\, kg$ is suspended from a fixed point by a wire of length $3\, m$ and diameter $0.5\, mm$. 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 ?
$\left(Y\right.$ of wire $\left.=206\, G P a, \alpha=11 \times 10^{-6}{ }^{\circ} C ^{-1}\right)$

Answer 1

Here, mass $=m=2\, kg$
Length of wire through which mass is suspended
$=\ell=3\, m$
Diameter of wire $=d=5 \times 10^{-4} m$
$Y_{\text {wire }}=200\, GPa =200 \times 10^{9} N / m ^{2}$
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, $\Delta \ell_{\text {wire }}=\ell \alpha \Delta T$
$1.454 \times 10^{-3} =3 \times 10^{-6} \times 11 \times \Delta T$
$\Delta T =\frac{1.454 \times 10^{-3}}{33 \times 10^{-6}}$
$=44.06^{\circ} C$