__Question 9.5__:Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

__Solution__:^{–4}m

Elongation of the brass wire = 1.3 × 10

^{–4}m

Diameter of the wires,

*d*= 0.25 m

Hence, the radius of the wires, r = d/2 = 0.125 cm

Length of the steel wire,

*L*

_{1}= 1.5 m

Length of the brass wire,

*L*

_{2}= 1.0 m

Total force exerted on the steel wire:

*F*

_{1}= (4 + 6) g = 10 × 9.8 = 98 N

Young’s modulus for steel:

Y

_{1}= (F

_{1}/A

_{1}) / (Δ

*L*

_{1}/ L

_{1})

Where,

Δ

*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

∴ Δ

*L*

_{1}= F

_{1}X L

_{1}/ (A

_{1}X Y

_{1})

= (98 X 1.5) / [ π(0.125 X 10

^{-2})

^{2}X 2 X 10

^{11}] = 1.49 X 10

^{-4}m

Total force on the brass wire:

*F*

_{2}= 6 × 9.8 = 58.8 N

Young’s modulus for brass:

Y

_{2}= (F

_{2}/A

_{2}) / (Δ

*L*

_{2}/ L

_{2})

Where,

Δ

*L*

_{2}= Change in the length of the brass wire

*A*

_{1}= Area of cross-section of the brass wire = πr

_{1}

^{2}

∴ Δ

*L*

_{2}= F

_{2}X L

_{2}/ (A

_{2}X Y

_{2})

= (58.8 X 1) / [ (π X (0.125 X 10

^{-2})

^{2}X (0.91 X 10

^{11}) ] = 1.3 X 10

^{-4 }m

Elongation of the steel wire = 1.49 × 10

^{–4}m

Elongation of the brass wire = 1.3 × 10

^{–4}m

## No comments:

## Post a Comment