A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the wire is l.
A
Loss in gravitational potential energy of M is Mgl
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B
The elastic potential energy stored in the wire is Mgl
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C
The elastic potential energy stored in the wire is 12Mgl
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D
Heat produced is 12Mgl
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Solution
The correct options are A Heat produced is 12Mgl C Loss in gravitational potential energy of M is Mgl D The elastic potential energy stored in the wire is 12Mgl We know that gravitational potential energy is VG=Mgh , where h= distance from equilibrium position. so here h=l, loss in gravitational P.E. is VG=Mgl
Elastic potential energy is Ve=12Fl=12Mgl
As the energy is conserved so loss in P.E. is equal to the sum of elastic potential energy stored and heat energy.