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11 months ago

Arun
25768 Points
```							The work done to lift the hanging part of the cable up to the surface is MgL/2n²Given-Mass of uniform cable = MLength of cable = LLength of cable hanging below the edge of the surface is = 1/n th partMass of the hanging part = M/nCenter of mass of h = L/2nSo work done should be-W = mgh = (M/n) g (L/2n)W = MgL/2n²
```
11 months ago
Vikas TU
14146 Points
```							Dear student There is short cut.Work done against gravity = mgh = m/n g l/2n = mgl /2n^2Good Luck Cheers
```
11 months ago
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### Course Features

• 101 Video Lectures
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions