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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)

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)

$nMgL$

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