Question

Two similar wires are stretched by the same load and the difference in their elongation is 0.25 cm. if their elasticities are in the ratio 5:3, find their individual elongations.​

Answer 1

Explanation:

Diameter of the wires, d =0.25m

Hence, the radius of the wires, r=

2

d

=0.125cm

Length of the steel wire, L

1

=1.5m

Length of the brass wire, L

2

=1.0m

Total force exerted on the steel wire:

F

1

=(4+6)g=10×9.8=98N

Young’s modulus for steel:

Y

1

=

(

L

1

△L

1

)

(

A

1

F

1

)

Where,

△L

1

= Change in the length of the steel wire

A

1

= Area of cross-section of the steel wire =πr

1

2

Young’s modulus of steel, Y

1

=2.0×10

11

Pa

∴△L

1

=

(A

1

×Y

1

)

F

1

×L

1

=(98×1.5)/[π(0.125×10

−2

)

2

×2×10

11

]=1.49×10

−4

m

Total force on the brass wire:

F

2

=6×9.8=58.8N

Young’s modulus for brass:

Y

2

=

(

L

2

△L

2

)

(

A

2

F

2

)

Where,

△L

2

= Change in the length of the brass wire

A

2

= Area of cross-section of the brass wire =πr

1

2

∴△L

2

=

(A

2

×Y

2

)

F

2

×L

2

=(58.8×1)/[(π×(0.125×10

−2

)

2

×(0.91×10

11

)]=1.3×10

−4

m

Elongation of the steel wire =1.49×10

–4

m

Elongation of the brass wire =1.3×10

−4

m.

Solve any question of Mechanical

Answer 2

Answer:

Individual elongations are 0.625cm and 0.375cm.

Explanation:

Young's modulous of elasticity is given by = $Y= \frac{stress}{strain} = \frac{F \Delta L}{AL}$

where, $F$ is the force acting on the wire,

$\Delta L$ is the elongation,

$L$ is the length of the wire,

$A$ is the cross-section area of the area..

Let the elasticities of the two wires be $Y_1, Y_2$ respectively.

and thier respective elongations be $\Delta L_1,\Delta L_2.$

From the question $F, L, A$ are constant.

Therefore, $Y_1=\frac{F (\Delta L_1)}{A.L}$

and $Y_2=\frac{F (\Delta L_2)}{A.L}$

Given: $\Delta L_2-\Delta L_1=0.25cm$ -----(1)

   $\frac{Y_1}{Y_2}=\frac{5}{3}$

$\implies \frac{Y_1}{Y_2}=\frac{\frac{F (\Delta L_1)}{A.L}}{\frac{F (\Delta L_2)}{A.L}} = \frac{5}{3}$

$\implies \frac{\Delta L_1}{\Delta L_2}=\frac{5}{3}$

$\implies \Delta L_1=\frac{5}{3}\Delta L_2$

Putting this in equation (1),

$\frac{5}{3}\Delta L_2-\Delta L_2 =0.25$

$\implies \frac{2}{3}\Delta L_2 =0.25$

$\implies \Delta L_2 =0.25 *\frac{3}{2}$

$\implies \Delta L_2 =0.375cm$

$\implies \Delta L_1=\frac{5}{3}\Delta L_2 = \frac{5}{3}*0.375$

$\implies \Delta L_2 =0.625cm$

Individual elongations are 0.625cm and 0.375cm.

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