A brass wire 1.8 m long at 27 °C is held taut with little tension bet

1.	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.
Answer» *Ur Answer* :-** Initial temperature, T = 27°C Length of the brass wire at T,L = 1.8m Final temperature, T2 = 39C Diameter of the wire, d = 2.0 mm = 2x10m Tension developed in the wire =F Coefficient of linear EXPANSION of brass, = 2.0x 10SK-1 Youngs modulus of brass, Y = 0.91 x 10 Pa Youngs modulus is given by the relation: Y= Stress/ Strain y = F/A tri.L/L AL/L AL = F x L/(A x Y) .) Where, F = Tension developed in the wire A Area of cross-section of the wire. AL =Charnge in the length. given by the relation: AL = aL (T, -Ti)..ii) Equating equations (i) and (ii). we get: FL aL = a (T²-T¹) = FL/ π (d /2) ² Y F a(T2 -T1)Y T(d/2 F =2x 10 x (-39 27) x 3.14 x 0.91 x 10 x (2 x 10-s/2 F =-3.8 x 10 N (The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is 3.8 x10 N.

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.

Answer»

Initial temperature, T = 27°C

Length of the brass wire at T,L = 1.8m

Final temperature, T2 = 39C

Diameter of the wire, d = 2.0 mm = 2x10m

Tension developed in the wire =F

Coefficient of linear EXPANSION of brass, =

2.0x 10SK-1

Youngs modulus of brass, Y = 0.91 x 10 Pa

Youngs modulus is given by the relation:

y = F/A

tri.L/L

AL/L

AL = F x L/(A x Y) .)

Where,

F = Tension developed in the wire

A Area of cross-section of the wire.

AL =Charnge in the length. given by the

relation:

AL = aL (T, -Ti)..ii)

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

aL = a (T²-T¹) = FL/ π (d /2) ² Y

F a(T2 -T1)Y T(d/2

F =2x 10 x (-39 27) x 3.14 x 0.91 x

10 x (2 x 10-s/2

F =-3.8 x 10 N

(The negative sign indicates that the tension

is directed inward.)

Hence, the tension developed in the wire is

3.8 x10 N.