The young modules of an wire of length ‘L’ and cross sectional area `alpha` is y The wire is extended in length within elastic‘limit with a stress's’. Calculate the energy density stored in the wire.

Initial length of a wire is L and its area of cross section is A. The wire is elongated by applying a stress tau within its elastic limit. Prove that the potential energy density in the wire due to elongation is (tau^2)/(2Y) [Y=Young's modulus]

A wire of initial length L and area of cross section A has Young's modulus Y of its material. The wire is stretched by a stress 5 within its elastic limit. The stored energy density in the wire will be

A wire of length L and cross sectional area a is stretched by l, where l lt L . Show that if the elastic limit is not exceeded, the potential energy of the wire increases by (Yal^2)/(2L)

Two conducting wires of lenghts l and 2l have the same cross-sectional area.Compare their resistances.

A load of 20 kg is hung from one end of a wire of length 6 m and of cross sectional area 1mm^2 . IF the load is withdrawn, the length of the wire becomes 5.995 m. Calculate the (i) longitudinal strain, (ii) longitudinal stress and (iii) Young's modulus for the material of the wire.

A wire of length L and cross section A hung from a rigid support is loaded with a mass M The elongation produced is

For a uniform wire of length 3 m and cross sectional area 1mm^2 , 0.021 J of work is necessary to stretch it through 1mm. Calculate the Young's modulus for its material.

Two wires A and B are made of the same metal. Diameter of A is double that of B and the length of A is thrice that of B. IF both the wires are stretched by the same force to elongate them equally within elastic limit, then the ratio of energy stored in the wires A and B will be