A body of mass M is attached to lower end of a metal wire, whose upper end is fixed. The elongation of wire is l.
A
Loss in gravitational potential energy of M is Mgl.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The elastic potential energy stored in the wire is Mgl.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
The elastic potential energy stored in the wire is 12Mgl.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Heat produced is 12Mgl
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are A The elastic potential energy stored in the wire is 12Mgl. B Heat produced is 12Mgl C Loss in gravitational potential energy of M is Mgl.
Since it moves by l, gravitational potential energy will be Mgl.
Elastic potential energy = 1/2×Stress×Strain×Volume=1/2×MgA×lL×AL=Mgl2
Heat produced =Change in energy=Elastic energy stored-Gravitational energy lost=−Mgl2. Thus heat lost=Mgl2