Question

A force F produces an elongation l, in a wire. what force is required to produce the same elongation in the same material and radius, but of length twice the first one? ​

Answer 1

___First Wire___

Force = F

Area = a

Length = l

Elongation in length = ∆l

Young's Modulus of first material = ( F / a ) × ( l / ∆l )

___Second Wire___

Force = f

Area = A

Length = L

Elongation in length = ∆L

Young's Modulus of second material = ( f / A ) × ( L / ∆L )

Both wires are made up of same material.

•°•

Young's Modulus of first material = Young's Modulus of second material

( F / a ) × ( l / ∆l ) = ( f / A ) × ( L / ∆L )

Same elongation and same radius

So, F × l = f × L

Now, the length of second wire is double the length of first wire.

So, L = 2l

•°•

F × l = f × 2l

F = 2f

f = 2 / F

Hence, Half of the force applied on first wire should be applied on the second wire.