Ncert Exercises & Solutions

11.9 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? Co-efficient of linear expansion of brass = $2.0 \times 10^{- 5} K^{- 1}$ ; Young’s modulus of brass = $0.91 \times 10^{11} Pa$ .

Initial temperature, T $_{1}$ = 27°C

Length of the brass wire at T $_{1}$ , l = 1.8 m

Final temperature, T $_{2}$ = –39°C

Diameter of the wire, d = 2.0 mm = 2 × $10^{- 3}$ m

Young’s modulus is given by:

$Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{\frac{∆ L}{L}} ∆ L = \frac{F \times L}{A \times Y} . . . (i)$

where,

F = Tension developed in the wire

A = Area of the cross-section of the wire

ΔL = Change in the length= Lα(T2 – T1) … (ii)

$\Rightarrow αL (T_{1} - T_{2}) = \frac{FL}{π {(\frac{d}{2})}^{2} \times Y} F = 2 \times 10^{- 5} (- 39 - 27) \times 3.14 \times 0.91 \times 10^{11} \times {(\frac{2 \times 10^{- 3}}{2})}^{2} = - 3.8 \times 10^{- 2} N$

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is 3.8 ×10 $^{2}$ N.

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