Question

Suppose a steel wire whose length is 130cm and whose diameter is 1.1mm is heated to an average temperature of 83°C and stretch out between two rigid supports.What tension develops in the wire as it cools to 20°C

Answer 1

Given that,

Length = 130 cm

Diameter = 1.1 mm

Temperature T₁= 20° C

Temperature T₂= 83° C

We need to calculate the change in length

Using formula of coefficient of linear expansion

$\alpha=\dfrac{\Delta L}{L\times\Delta T}$

$\Delta L=\alpha\times L\times\Delta T$

Put the value into the formula

$\Delta L=1.3\times10^{-5}\times1.3\times(20-83)$

$\Delta L=0.001065\ m$

$\Delta L=1.065\ mm$

The change in length is 1.065 mm.

We need to calculate the tension develops in the wire

Using formula of young's modulus

$Y=\dfrac{FL}{A\Delta L}$

$F=\dfrac{AY\Delta L}{L}$

Where, A = area

L = length

F = force

$\Delta L$ = change length

Put the value into the formula

$F=\dfrac{\pi\times(\dfrac{1.1\times10^{-3}}{2})^2\times2\times10^{11}\times0.001065}{130\times10^{-2}}$

$F=155.7\ N$

Hence, The tension develops in the wire is 155.7 N.