Question

A steel wire 2mm in diameter is stretched between 2 clamps when it's temp. is 40°c . Calculate the tension in the wire when it's temp. falls 30°c ? Given, Coefficient of linear expansion of steel = 11 × 10^ -6 ° c ^ -1 And Y (young's modulus ) of steel = 21 × 10 ^ 11 dyne/cm^2.​

Answer 1

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}} \\ \\ \huge\bold{Given:-} \\ \\ Radius = 1mm = 0.01 cm \\ \\ \Delta l = l \times \alpha \Delta t \\ \\ \Delta l = l \times 11 \times {10}^{-6} \times (40-30) \\ \frac{ \Delta l}{l} = \alpha \times \Delta t = {10}^{-5}$ $\\ \large\boxed{ \Sigma = {10}^{-5}}$ $\\ \frac{\sigma (Stress)}{\Sigma (Strain) } = Y (Young's Modulus) \\ \\ 21 \times {10}^{11} = \frac{\sigma}{{10}^{-5}} \\ \\ \sigma = 21 \times {10}^{6} \\ \\ Now, \: \: for \: \: Tension, \\ \\ Tension = \sigma (stress) \times Area \\ \\ Tension = \cancel{21} \times {10}^{6} \times \frac{22}{\cancel{7}} \times ({10}^{-4}) \\ \\ \huge\boxed{ Tension = 66 \times {10}^{2}}$

Answer 2

Given that,

Diameter of wire = 2 mm

Initial temperature = 40°C

Final temperature = 30°C

Coefficient of linear expansion$\alpha=11\times10^{-6}^{\circ}C^{-1}$

Young's modulus $Y=21\times10^{11}\ dyne/cm^{2}$

We need to calculate the strain

Using formula of strain

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

Put the value into the formula

$\Delta l=l\times11\times10^{-6}\times(40-30)$

$\dfrac{\Delta l}{l}=10^{-5}$

We need to calculate the stress

Using formula of young's modulus

$Y=\dfrac{\sigma}{\dfrac{\Delta l}{l}}$

$\sigma=Y\times\dfrac{\Delta l}{l}$

Put the value into the formula

$\sigma=21\times10^{11}\times10^{-5}$

$\sigma=21\times10^{6}$

We need to calculate the tension in the wire

Using formula of tension

$T=\sigma\times A$

Where, A = area

$\sigma$ = stress

Put the value into the formula

$T=21\times10^{6}\times\pi\times(10^{-4})$

$T=66\times10^{2}\ N$

Hence, The tension in the wire is $66\times10^{2}\ N$