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The stress required to double the length of a wire of Young's modulus $' E'$ is____

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$E$

$2 E$

$4 E$

$\frac{E}{2}$

answer is A.

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Detailed Solution

$l_{2} = 2 l, e = l, \frac{e}{l} = 1 \Rightarrow Y = \frac{L . s t r e s s}{L . s t r a i n} \Rightarrow L . s t r e s s = Y \times L . s t r a i n$

Strain $= 1$

Stress $= y = E$

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