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A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable to the surface, the work done should be:

1. MgL/(2n²)
2. MgL/n²
3. 2MgL/n²
4. nMgL

Correct answer: MgL/(2n²)

Solution

The work done to lift the hanging part of the cable is calculated by considering the weight of the hanging segment and the distance it needs to be raised. Since only a fraction (1/n) of the cable is hanging, the average height of this segment is (L/(2n)), leading to the total work done being MgL/(2n²).

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