__Question 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 ×
10

^{–5}K^{–1}; Young’s modulus of brass = 0.91 × 10^{11}Pa.

__Solution__:*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

Tension developed in the wire =

*F*

Coefficient of linear expansion of brass,

*α*= 2.0 × 10

^{–5}K

^{–1}

Young’s modulus of brass,

*Y*= 0.91 × 10

^{11}Pa

Young’s modulus is given by the relation:

γ = Stress / Strain = (F/A) / (∆L/L)

∆L = F X L / (A X Y) ......(i)

Where,

*F*= Tension developed in the wire

*A*= Area of cross-section of the wire.

Δ

*L*= Change in the length, given by the relation:

Δ

*L*=

*α*

*L*(

*T*

_{2}–

*T*

_{1}) … (

*ii*)

Equating equations (

*i*) and (

*ii*), we get:

αL(T

_{2}- T

_{1}) = FL / [ π(d/2)

^{2}X Y ]

F = α(T

_{2}- T

_{1})πY(d/2)

^{2}

^{}

F = 2 X 10

^{-5}X (-39-27) X 3.14 X 0.91 X 10

^{11}X (2 X 10

^{-3}/ 2 )

^{2}

= -3.8 X 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.

## No comments:

## Post a Comment