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Q.

To double the length of a iron wire having  0.5 cm2 area of cross-section, the required force will be (Y=1012 dyne/cm2)

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a

1.0×10−7N

b

1.0×107N

c

0.5×10−7N

d

0.5×1012 dyne

answer is D.

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Detailed Solution

If length of wire doubled then strain = 1Y= stress ⇒F=Y×A=1012×0.5=0.5×1012 dyne
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To double the length of a iron wire having  0.5 cm2 area of cross-section, the required force will be (Y=1012 dyne/cm2)