Question

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.

Answer 1

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 .

Answer 2

Answer:

3.8×10^2N

Explanation:

Initial temperature, T1=27oC

Length of the brass wire at T1,l=1.8m

Final temperature, T2=39C 

Diameter of the wire, d=2.0mm=2×10−3m

Tension developed in the wire =F

Coefficient of linear expansion of brass, =2.0×10−5K−1

Youngs modulus of brass, Y=0.91×1011 Pa

Youngs modulus is given by the relation:

Y= Stress / Strain

Y=ΔL/LF/A

ΔL=F×L/(A×Y) ......(i)

Where,

F= Tension developed in the wire 

A= Area of cross-section of the wire. 

ΔL= Change in the length, given by the relation:

ΔL= αL(T2 -T1) .....(ii)

Equating equations (i) and (ii), we get:

αL(T2−T1)=π(d/2)2YFL

F=α(T2−T1)Yπ(d/2)2

F=2×10−5×(−39−27)×3.14×0.91×1011×(2×10−3/2)2

F=−3.8×102N

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

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