Ncert Exercises & Solutions

9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of Young’s modulus of steel to that of copper?

Given:

Length of the steel wire, L $_{1}$ = 4.7 m

Area of the cross-section of the steel wire, A $_{1}$ = 3.0 × 10-5 m2

Length of the copper wire, L $_{2}$ = 3.5 m

Area of the cross-section of the copper wire, A $_{2}$ = 4.0 × 10-5 m2

Change in length in both cases is same; = ΔL $_{1}$ = ΔL $_{2}$ = ΔL.

Force applied in both the cases is also same.

For the steel wire:

$Y_{1} = \frac{F_{1}}{A_{1}} \times \frac{L_{1}}{∆ L} = \frac{F \times 4.7}{3.0 \times 10^{- 5} \times ∆ L}$

For the copper wire:

$Y_{2} = \frac{F_{2}}{A_{2}} \times \frac{L_{2}}{∆ L} = \frac{F \times 4.7}{4.0 \times 10^{- 5} \times ∆ L}$

$\Rightarrow \frac{Y_{1}}{Y_{2}} = \frac{4.7 \times 4.0 \times 10^{- 5}}{3.0 \times 10^{- 5} \times 3.5} = 1.79 : 1$

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

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