Question

A tungsten wire, $0.5\,{\text{mm}}$ in diameter is just stretched between two fixed points at a temperature $40^\circ {\text{C}}$ . Determine the tension in the wire when the temperature falls to ${\text{20}}^\circ {\text{C}}$ . (Coefficient of linear expansion of tungsten is ${\text{4}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 6}}^\circ {{\text{C}}^{ - 1}}$ ; Young’s modulus of tungsten is ${\text{3}}{\text{.45}} \times {\text{1}}{{\text{0}}^{10}}\,{\text{N}}{{\text{m}}^{ - 2}}$ )
A. $0.6097\,{\text{N}}$
B. $0.3097\,{\text{N}}$
C. $0.5097\,{\text{N}}$
D. $0.7097\,{\text{N}}$

Answer 1

First of all, we will find the temperature difference and then will find the expression for Young’s modulus. Then we will rearrange the formula. After that we will find the expression for the change in length which is dependent on the coefficient of linear expansion of that material. Then we will substitute the required values in the main equation to find the tension. We will manipulate accordingly to obtain the result.

Complete step by step answer: In the given question, we are supplied the following data:
The diameter of the tungsten wire is $0.5\,{\text{mm}}$ .
The initial temperature is $40^\circ {\text{C}}$ and the final temperature is ${\text{20}}^\circ {\text{C}}$ .
Young’s modulus $\left( \gamma \right)$ of tungsten is ${\text{3}}{\text{.45}} \times {\text{1}}{{\text{0}}^{10}}\,{\text{N}}{{\text{m}}^{ - 2}}$ .
Coefficient of linear expansion $\left( \alpha \right)$ of tungsten is ${\text{4}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 6}}^\circ {{\text{C}}^{ - 1}}$ .

To begin with, we will find the temperature difference in the following:
${T_1} = 40^\circ {\text{C}}$
${T_2} = 20^\circ {\text{C}}$

$$\Delta T = {T_2} - {T_1} \\\ \Rightarrow \Delta T = 40^\circ {\text{C}} - 20^\circ {\text{C}} \\\ \Rightarrow \Delta T = 20^\circ {\text{C}} \\\$$

Now, we have a formula which gives Young’s modulus:
$\gamma = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}$
$\dfrac{F}{A} = \gamma \times \dfrac{{\Delta l}}{l}$ …… (1)
Where,
$F$ indicates the tension force in the string.
$A$ indicates the cross-sectional area of the wire.
$\gamma$ indicates the Young’s modulus of tungsten.
$\Delta l$ indicates the change in length.
$l$ indicates the original length of the wire.

We have a formula which gives the thermal expansion in length:
$\Delta l = l \times \alpha \times \Delta {\rm T}$ …… (2)
Where,
$\Delta l$ indicates the change in length.
$\alpha$ indicates coefficient of linear expansion of tungsten.
$\Delta {\rm T}$ indicates the change in temperature.
$l$ indicates the original length of the wire.

From equation (2), we have:
$\dfrac{{\Delta l}}{l} = \alpha \times \Delta {\rm T}$ …… (3)

Since, the diameter is given as $0.5\,{\text{mm}}$ .
We will convert the millimetres into metres.

$$0.5\,{\text{mm}} \\\ \Rightarrow 0.5 \times {10^{ - 3}}\,{\text{m}} \\\$$

Using equation (3) in equation (1), we get:

$$F = A \times \gamma \times \dfrac{{\Delta l}}{l} \\\ \Rightarrow F = \pi \times {\left( {\dfrac{{0.5 \times {{10}^{ - 3}}}}{2}} \right)^2} \times {\text{3}}{\text{.45}} \times {\text{1}}{{\text{0}}^{10}} \times \alpha \times \Delta {\rm T} \\\ \Rightarrow F = 6.77 \times {\text{1}}{{\text{0}}^3} \times {\text{4}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 6}} \times 20 \\\ \Rightarrow F = 0.6093\,{\text{N}} \\\$$

$\Rightarrow F \approx 0.6097\,{\text{N}}$
Hence, the tension in the wire when the temperature falls to ${\text{20}}^\circ {\text{C}}$ is $0.6097\,{\text{N}}$ .
The correct option is A.

Note: While solving this problem, be sure that all the units are in S.I units. The change in length of the material is directly dependent on the change in temperature. Higher the strain is, higher is the stress.