First slide

Thermal expansion

Question

A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x $10^{- 5} /^{0} C$ , Young's modulus of brass = 0.91 x $10^{11}$ Pa.

Moderate

A

$2.4 \times 10^{2} N$
B

$3.8 \times 10^{2} N$
C

$1.8 \times 10^{2} N$
D

$4.8 \times 10^{2} N$

Solution

As the wire is not free to contract, the thermal stress is developed at the wire.
The change in temperatwe,
$∆ T = 27^{0} C - (- 39^{0} C) = 66^{0} C$
Let $∆$ L be the change in length of the wire.
$∆ L = αL ∆ T = (2 \times 10^{- 5}) \times 1.8 \times 66 = 2.376 \times 10^{- 3} m$
As $Y = \frac{FL}{A ∆ L} = \frac{FL}{\frac{({πD}^{2})}{4} ∆ L} = \frac{4 FL}{{πD}^{2} ∆ L}$
F(tension in the wire) = $\frac{{YπD}^{2} ∆ L}{4 L}$
$= \frac{(0.91 \times 10^{11}) \times 3.142 \times {(2 \times 10^{- 3})}^{2} \times (2.376 \times 10^{- 3})}{4 \times 1.8}$
= $3.8 \times 10^{2} N$

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