InterviewSolution
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InterviewSolution
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A rod of mass `m` and length `l` is lying on a horizontal table. Work done in making it stand on one end will beA. `mgl`B. `mgl//2`C. `mgl//4`D. `2mgl` |
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Answer» (b) Work done = final PE. - initial PE. = `mg (1//2) - 0 = mgl//2` (as when the rod is standing on one end, its centre of gravity is at a distance `1//2` from the table and when it is lying on the table, its CG is on the table, i.e., at zero distance from the table) |
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