Question

A tungsten wire, $0.5mm$ in diameter is just stretched between two fixed point at a temperature ${40}^o$C. Determine the tension in the wire when the temperature falls to ${20}^{o}C$.(coefficient of linear expansion of tungsten = $4.5\times {10}^{-6}{/}^{o}C$;Young's modulus of tungsten $=3.45\times {10}^{10}N{m}^{-2}$)

• A. $0.6097N$
• B. $0.3097N$
• C. $0.5097N$
• D. $0.7097N$

Answer 1

The correct option is A $0.6097N$

Given,

$\gamma =3.45\times {10}^{10}N{m}^{-2}$

$\alpha =4.5\times {10}^{-6}{/}^{0}C$

$D=0.5mm$

$T_1={40}^{0}C$

$T_2={20}^{0}C$

$\Delta T={20}^{0}C$

Young modulus, $\gamma =\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$

$\frac{F}{A}=\gamma \frac{\Delta l}{l}$. . . .(1)

The thermal expansion in length,

$\Delta l=l\alpha \Delta T$

$\frac{\Delta l}{l}=\alpha \Delta T$. . . .(2)

Area $A=\frac{\pi D^2}{4}$. . . .(3)

Put equation (2) and (3) in equation (1), we get

$F=\frac{\pi D^2}{4}\gamma \alpha \Delta T$

$F=\frac{3.14\times 0.5\times 0.5\times {10}^{-6}}{4}\times 3.45\times {10}^{10}\times 4.5\times {10}^{-6}\times 20$

$F=0.6093N$