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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

A uniform cable of mass \(‘M’\) and length \(‘L’\) is placed on a horizontal surface such that its \({\left( {\frac{1}{n}} \right)^{th}}\)  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

  1. A \(nMgL\)
  2. B \(\frac {MgL}{2n^2}\)
  3. C \(\frac {2MgL}{n^2}\)
  4. D \(\frac {MgL}{n^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac {MgL}{2n^2}\)

Step-by-step Solution

Detailed explanation

Mass of the hanging part \(=\) \(\frac{M}{n}\) \({h_{COM}} = \frac{L}{{2n}}\) work done \(W = mg{h_{COM}} = \left( {\frac{M}{n}} \right)g\left( {\frac{L}{{2n}}} \right) = \frac{{MgL}}{{2{n^2}}}\)
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