The force required to stretch a steel wire of 1 cm² cross-section to 1.1 times its length would be: $(Y = 2 \times 10^{11} N m^{- 2})$

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$2 \times 10^{3} N$

$2 \times 10^{6} N$

$2 \times 10^{- 7} N$

$2 \times 10^{- 6} N$

answer is A.

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Detailed Solution

$L_{2} = 1.1 L_{1} ∴ S t r a i n = \frac{l}{L_{1}} = \frac{L_{2} - L_{1}}{L_{1}} = \frac{1.1 L_{1} - L_{1}}{L_{1}} = 0.1 \Rightarrow F = Y A \frac{l}{L} = 2 \times 10^{11} \times 1 \times 10^{- 4} \times 0.1 = 2 \times 10^{6} N$

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