### Q.1:- A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^{–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?

**Answer:-**Length of the steel wire, *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}/ Δ

*L*

_{2})

= (

*F*/ 4 × 10

^{-5}) (3.5 / Δ

*L)*….(ii)

Dividing (

*i*) by (

*ii*), we get:

*Y*

_{1}/

*Y*

_{2}= (4.7 × 4 × 10

^{-5}) / (3 × 10

^{-5}× 3.5)

= 1.79 : 1

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