JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2019
- Hard
- MCQ
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
Pick an option to check your answer.
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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