A cylindrical rod of circular cross section is subjected to a tensile force and the longitudinal strain produced in the rod is 2%. If the radius of the rod is doubled and the applied force is also doubled, the longitudinal strain produced will be
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answer is 4.
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Detailed Solution
e=∆ll=FA.Y=Fπr2Y⇒e2e1=r1r22F2F1=122x 2=12⇒e2=12 x 2%=1%