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 3.5m 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

0.9

1.8

3.6

answer is A.

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Detailed Solution

$\large {\ell _{st}} = 4.7m\,\,{\ell _{cu}} = 3.5m;\;{A_{st}} = 3 \times {10^{ - 5}}{m^2}$
$\large {A_{cu}} = 4 \times {10^{ - 5}}{m^2};\frac{{{y_{st}}}}{{{y_{cu}}}} = ?$
F is same e is same $\large y = \frac{F}{A} \times \frac{\ell }{e} \Rightarrow y\alpha \frac{\ell }{A}$
$\large \frac{{{y_{st}}}}{{{y_{cu}}}} = \frac{{{\ell _{st}} \times {A_{cu}}}}{{{\ell _{cu}} \times {A_{st}}}} = \frac{{4.7}}{{3.5}} \times \frac{{4 \times {{10}^{ - 5}}}}{{3 \times {{10}^{ - 5}}}} = \frac{{188}}{{105}} = 1.8$