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`B. The elastic potential energy stored in the wire is ` Mgl`C. The elastic potential energy stored in the wire is `1/2 Mgl`D. Heat produced is `1/2 Mgl`

A body of mass `M` is attached to the lower end of a metal wire, whose

1.	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`B. The elastic potential energy stored in the wire is ` Mgl`C. The elastic potential energy stored in the wire is `1/2 Mgl`D. Heat produced is `1/2 Mgl`
Answer» Correct Answer - A::C::D Half of energy is lost in heat and rest half is stored as elastic potential energy . ` l = (Mgl)/(AY)` …(i) ` U = 1/2 Kl^(2) = 1/2((YA)/(L))l^(2)` …(ii) From Eqs. (i) and (ii) , we can prove that `U = 1/2 Mgl`

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