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Two wires of the same length and radius are stretched by the same load and the difference of their elongations produced is found to be 0.25 cm . If their modulus of elasticity are in the ratio 5:3. Then the individual elongations are

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a

0.375cm, 0.625 cm

b

0.625 cm, 0.650 cm

c

0.325 cm, 0.350cm

d

0.675 cm, 0.325cm

answer is A.

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Detailed Solution

e2−e1=0.25cm∵Y1>Y2 Y=FlAe⇒e∝1y∵Y,F,l are same⇒e1e2=y2y1=35e1=35e2 ⇒e1=35e1+0.25⇒5e1=3e1+0.75 ⇒2e1=0.75⇒e1=0.375cm ⇒e2=e1+0.25= 0.375+0.25=0.625cm

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