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 × 10^11 Pa.​

Answer 1

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.

Answer 2

Given Information,

• Initial Temperature = 27° C
• Final Temperature = -39°C
• Length of the wire = 1.8m
• Diameter of the wire = 2mm
• Coefficient of linear expansion : 2 × 10^{-5} K^{-1}
• Young's Modulus = 0.91 × 10¹¹ Pa

We know that,

$\sf Young's \ Modulus = \dfrac{Stress}{Strain}$

Now,

$\sf Y = \dfrac{\dfrac{F}{A}}{\dfrac{e}{L}} \\ \\ \implies \sf e = \dfrac{FL}{AY}---------(1)$

Also,

$\sf \: e = \alpha L \Delta{T} - - - - - - - (2)$

Equating (1) and (2), we get :

$\sf \dfrac{F \times L}{A \times Y} = \alpha L \Delta{T}$

Here,

• change in temperature is -66°C
• area of cross section is πr² = π(d/2)²

Substituting values, we get :

$\boxed{ \boxed{ \sf F= - 3.8 \times {10}^{2} \: N}}$