__Question 9.1__:^{–5}m

^{2}stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10

^{–5}m

^{2}under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

__Solution__:*L*

_{1}= 4.7 m

Area of cross-section of the steel wire,

*A*

_{1}= 3.0 × 10

^{–5}m

^{2}

Length of the copper wire,

*L*

_{2}= 3.5 m

Area of cross-section of the copper wire,

*A*

_{2}= 4.0 × 10

^{–5}m

^{2}

Change in length = Δ

*L*

_{1}= Δ

*L*

_{2}= Δ

*L*

Force applied in both the cases =

*F*

Young’s modulus of the steel wire:

Y

_{1}= (F

_{1}/ A

_{1}) (L

_{1}/ Δ

*L*

_{1})= (F / 3 X 10

^{-5}) (4.7 / Δ

*L)*....(i)

Young’s modulus of the copper wire:

Y

_{2}= (F

_{2}/ A

_{2}) (L

_{2}/ ΔL

_{2})

= (F / 4 X 10

^{-5}) (3.5 / Δ

*L)*....(ii)

Dividing (

*i*) by (

*ii*), we get:

Y

_{1}/ Y

_{2}= (4.7 X 4 X 10

^{-5}) / (3 X 10

^{-5}X 3.5)

= 1.79 : 1

The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

## No comments:

## Post a Comment