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.

Ajay Kumar Shukla , Meritnation Expert added an answer, on 15/6/10

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

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:

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:

(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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