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
- A \(nMgL\)
- B \(\frac {MgL}{2n^2}\)
- C \(\frac {2MgL}{n^2}\)
- D \(\frac {MgL}{n^2}\)
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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