JEE Main20199 Apr 2019Morning ShiftPhysicsWork, Power and EnergyActual
A uniform cable of mass M and length L is placed on a horizontal surface such that its 1 n t h part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:
Options
- AM g L 2 n 2
- BM g L n 2
- Cn M g L
- D2 M g L n 2
Correct answer
A. M g L 2 n 2
Step-by-step solution
Mass of the part hanging is M n . Centre of mass of this part lies L 2 n distance from the surface. ∴ Work done in lifting the hanging part to surface is W = M n · g L 2 n = M g L 2 n 2