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A metal wire of length L, area of cross-...

A metal wire of length L, area of cross-section A and young's modulus `Y` is stretched by a variable force `F` such that `F` is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is `l`

A

the work done by `F` is `(YAl^(2))/(2L)`

B

the work done by `F` is `(YAl^(2))/L`

C

the elastic potential energy stored in wire is `(YAl^(2))/(2L)`

D

no energy is lost during elongation

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
In the question the applied force is not constant. Let us consider at any instant the wire elongates by `x`. Then work doen by `F` to elongate it more by `dx` is
`dW=Fdx=(Yax)/Ldx[x lt lt L]`
`W=int_(0)^(1)(Yaxdx)/L=(Yal^(2))/(2L)`
As internal restoring force and applied force are almost same (equal and opposite) numerical value of work done by these two forces would be equal.
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