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Q.

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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a

0.45

b

1.8

c

3.6

d

0.9

answer is A.

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Detailed Solution

$l_{st} = 4.7 m l_{cu} = 3.5 m; A_{st} = 3 x 10^{- 5} m^{2} A_{cu} = 4 x 10^{- 5} m^{2}; \frac{Y_{st}}{Y_{cu}} = ?$
F is same e is same $Y = \frac{F}{A} x \frac{l}{e} \Rightarrow Y \propto \frac{l}{A}$
$\frac{Y_{st}}{Y_{cu}} = \frac{l_{st} x A_{cu}}{l_{cu} x A_{st}} = \frac{4.7}{3.5} x \frac{4 x 10^{- 5}}{3 x 10^{- 5}} = \frac{188}{105} = 1.8$

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