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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 × 1011 Pa.

Asked by Siddharth Arora(student) , on 15/6/10

Answers

EXPERT ANSWER

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.

Posted by Ajay Shukla(MeritNation Expert)on 15/6/10

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