Question

A brass wire 1.8 m long at 27^(@)C is held taut with little tension between two rigid supports. If the wire 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 xx 10^(-5)//^(@)C, Young's modulus of brass = 0.91 xx 10^(11) Pa.

Answer 1

Answer:

6gb5cpd5cogf6d7d0f57f9f5fpf55v0f6d5

Answer 2

Answer:

-377N

Explanation:

Concept :- when a wire or rod held taut between two rigid support is cooled then , stress is produced in it , called thermal stress.

Thermal stress ( F/A) = Young's modulus × strain

= Y∆L/L

= Ya∆T

Where a is coefficient of linear expansion of wire and ∆T temperature change .

Here,

Length of wire = 1.8 m

Intial temperature= 27°C

final temperature = -39°C

Temperature change ( ∆T) = -39-27 = -66°C

Diameter of wire ( d) = 2mm = 2 × 10^-3 m

coefficient of linear expansion ( a) = 2 × 10^-5 /K

Young's modulus ( Y) = 0.91 × 10¹¹ N/m²

So, tension developed in wire ( F) = YAa∆T

= Y( πd²/4)a.∆T

= 0.91 × 10¹¹ × 3.14 × (2 × 10^-3)² × 2 ×10^-5 × (-66)/4

= 0.91 × 3.14 × (-66)

= -377N

Here, negative sign shows that force is acting inwards due to contraction of the wire .

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