Question

A uniform wire of length L and radius r is twisted
by an angle a. If modulus of rigidity of the wire is
n. then the elastic potential energy stored in wire....​

Answer 1

The elastic potential energy stored in wire is  P.E = 1/2Ux²(πR²l)

Explanation:

We are given:

• Length of the wire = l
• Radius of the wire = r

Thus,

Shear modulus U = The shear stress / The shear strain

Shear stress = U×shear strain

Work done per unit volume will be = 1/ 2 × stress × strain

Work done = 1/2 Ux²

P.E = 1/2Ux²(πR²l)

A wire can be considered to the cylinder.

Hence the elastic potential energy stored in wire is  P.E = 1/2Ux²(πR²l)

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