# A uniform cable of mass $'M'$ and length $'L'$ is placed on a horizontal surface that its $\inline \left [ \frac{1}{n} \right ]^{th}$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)
S solutionqc

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)

$nMgL$

Exams
Articles
Questions