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A steel wire of length 4.7 m and cross - sectional area 3×10^–5m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4×10^–5m² under a given load. Then the ratio of the Young's modulus of steel to that of copper

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0.45

1.8

3.6

0.9

answer is A.

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Detailed Solution

In case of steel,

$I_{1} = 4.7 m A_{1} = 3.0 \times 10^{- 5} m^{2}$

$Y_{1} = \frac{F \times 4.7}{3.0 \times 10^{5} \times ∆ L}______(1)$

For copper,

$l_{2} = 3.5 m A_{2} = 4.0 \times 10^{- 5} m^{2}$

$Y_{2} = \frac{F \times 3.5}{4.0 \times 10^{- 5} \times ∆ L}_______(2)$

Dividing equation 1 by equation 2

$\frac{Y_{1}}{Y_{2}} = \frac{\frac{F \times 4.7}{3.0 \times 10^{- 5} \times ∆ L}}{\frac{F \times 3.5}{4.0 \times 10^{- 5} \times ∆ L}}$

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

$\frac{Y_{1}}{Y_{2}} = \frac{4.7 \times 4.0}{3.0 \times 3.5} \frac{Y_{1}}{Y_{2}} = 1.8$

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