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A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface 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 srface, the work done should be :

  • Option 1)

      \frac{MgL}{2n^{2}}

  • Option 2)

      \frac{MgL}{n^{2}}

  • Option 3)

     \frac{2MgL}{n^{2}}

  • Option 4)

     nMgL

 

Answers (1)

best_answer

Total length of rod=L

Length hanging=\frac{L}{n}

initial P.E = u_i=\left ( \frac{M}{n} \right )\left ( \frac{L}{n}\right )\frac{1}{2}g=\frac{MLg}{2n^{2}}

final P.E=u_f=\left ( \frac{M}{n} \right )g\left ( \frac{L}{n} \right )=\frac{MgL}{n^{2}}

               W=u_{f}-u_{i}

                   =\frac{MgL}{2n^{2}}

                


Option 1)

  \frac{MgL}{2n^{2}}

Option 2)

  \frac{MgL}{n^{2}}

Option 3)

 \frac{2MgL}{n^{2}}

Option 4)

 nMgL

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