Solved Numerical Problem No. 17.6 ( Unit-17: Physics of Solids); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

A cylindrical copper wire and a cylindrical steel wire each of length 1.5 m and diameter 2.0 mm is joined at one end to form a composite wire 3.0 m long. The wire is loaded until its length becomes 3.003 m. Calculate the strain in copper and steel wires and the force applied to the wire. (Young's modulus of copper is 1.2 ｘ 10¹¹ Pa and for steel is 2.0 ｘ 10¹¹ Pa). (Ans : 1.25 ｘ10⁻³, 7.5 ｘ 10⁻⁴, 471 N )

Data Given:

Length of copper wire =

L_{C}

$\L_C$ = 1.5 m

Length of steel wire =

L_{s}

$\L_s$ = 1.5 m

Total length = L = 1.5 m + 1.5 m = 3.0 m

Diameter of copper wire = Diameter of steel wire = D = 2.0 mm = 2.0 ｘ10⁻³ m

Radius = r =

\frac{D}{2}

$\frac {D}{2}$ =

$\frac {2.0 ｘ10⁻³ m }{2}$ = 1.0 ｘ10⁻³ m

Length of composite wire = L' = 3.003 m

Young's modulus of copper wire =

$\Y_C$ = 1.2 ｘ 10¹¹ Pa

Young's modulus of steel wire =

$\Y_S$ = 2.0 ｘ 10¹¹ Pa

From the given data we can calculate:

ΔL = L' - L = 3.003 m - 3.0 m = 0.003 m

and

Area = A = ㄫ r² = 3.1416 ｘ (1.0 ｘ10⁻³ m)² = 3.1416 ｘ10⁻⁶ m²

To Find:

Strain in copper wire =

$\Ɛ_C$ = ?

Strain in steel wire =

$\Ɛ_S$ = ?

Force = F = ?

Solution:

The formula for strain in copper wire is

$\Ɛ_C$ =

$\frac {ΔL_C}{L}$ ----------(1)

similarly formula for strain in steel wire

$\Ɛ_S$ =

$\frac {ΔL_s}{L}$ -------------(2)

Here we see that the value of

$\ΔL_C$ and

$\ΔL_S$ are unknown. So we have to find it first.

Let

The elongation in the copper wire =

$\ΔL_C$ then

The elongation in the steel wire =

$\ΔL_S$ = ΔL -

$\ΔL_C$

$\ΔL_S$ = 0.003 m -

$\ΔL_C$ --------(a)

Now using the formula of Young Modulus for copper and steel wire separately.

For Copper wire

$\Y_C$ =

$\frac {F}{A}$ ｘ

$\frac {L}{ΔL_C}$

Or

$\frac {F}{A}$ =

$\Y_C$ ｘ

$\frac {ΔL_C}{L}$ ---------(b)

For both copper and steel wire the magnitude of stress =

$\frac {F}{A}$ will be same So,

for steel wire

$\frac {F}{A}$ =

$\Y_S$ ｘ

$\frac {ΔL_S}{L}$ ---------(c)

By comparing eqn (b) and (c) we get

$\Y_C$ ｘ

$\frac {ΔL_C}{\cancel L}$ =

$\Y_S$ ｘ

$\frac {ΔL_S}{\cancel L}$

The quantity L will be canceled from both sides thus

$\Y_C$ ｘ

$\ΔL_C$ =

$\Y_S$ ｘ

$\ΔL_S$

By putting the value of

$\ΔL_S$ from equation (a)

$\Y_C$ ｘ

$\ΔL_C$ =

$\Y_S$ ｘ (0.003 m -

$\ΔL_C$ )

By putting all the corresponding values

1.2 ｘ 10¹¹ Pa ｘ

$\ΔL_C$ = 2.0 ｘ 10¹¹ Pa ｘ (0.003 m -

$\ΔL_C$ )

1.2 ｘ 10¹¹ Pa ｘ

$\ΔL_C$ = 0.006 ｘ 10¹¹ m Pa - 2.0 ｘ 10¹¹ Pa ｘ

$\ΔL_C$

1.2 ｘ 10¹¹ Pa ｘ

$\ΔL_C$ + 2.0 ｘ 10¹¹ Pa ｘ

$\ΔL_C$ = 0.006 ｘ 10¹¹ m Pa

3.2 ｘ 10¹¹ Pa ｘ

$\ΔL_C$ = 0.006 ｘ 10¹¹ m Pa

$\ΔL_C$ =

$\frac {0.006 ｘ 10¹¹ m Pa }{3.2 ｘ 10¹¹ Pa}$

$\ΔL_C$ = 0.001875 m --------------(d)

again using eqn (a)

$\ΔL_S$ = 0.003 m -

$\ΔL_C$ = 0.003 m - 0.001875 m

$\ΔL_S$ = 0.001125 m ------------(e)

Now for the calculation of strain in copper wire

Using Eqn (1)

$\Ɛ_C$ =

$\frac {ΔL_C}{L}$

$\Ɛ_C$ =

$\frac {0.001875 m}{1.5m}$

$\Ɛ_C$ = 0.00125 ----------Ans (1)

for calculation of strain in steel wire

Using Eqn (2)

$\Ɛ_S$ =

$\frac {ΔL_S}{L}$

$\Ɛ_S$ =

$\frac {0.001125 m}{1.5m}$

$\Ɛ_S$ = 0.00075 ----------Ans (2)

To the value of force applied to the wire, we have the formula of young modulus for:

Let for Copper wire

$\Y_C$ =

$\frac {F}{A}$ ｘ

$\frac {L}{ΔL_C}$

Or

$\frac {F}{A}$ =

$\Y_C$ ｘ

$\frac {ΔL_C}{L}$

F = A

$\Y_C$ ｘ

$\frac {ΔL_C}{L}$

By putting values

F = 3.1416 ｘ10⁻⁶ m² ｘ 1.2 ｘ 10¹¹ Pa ｘ

$\frac {0.001875 m}{1.5 m}$

F = 3.76992 ｘ10⁵ m² Pa ｘ 0.00125 m

[ 1 Pa = 1 N m⁻² ]

F = 0.0047124 ｘ10⁵ N m

F =471.24 N m -----------------Ans.3

Note: You can use the Young Modulus formula for steel wire with the same result for finding Fore "F"

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