A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its `(1/n)^(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:

Options

A. `(Mgl)/(2n^(2))`
B. `(Mgl)/(n^(2))`
C. `(2Mgl)/(n^(2))`
D. nMgl

Answer

b

Explanation

Length of banging part=`L//n`
Mass of hanging part = `M//n`
Weight of hanging part = `Mg//n`
Let .C. be the centre of mass of the hanging part.

The hanging part can be assumed to be a particle of weight `Mg//n` at a distance `L//n` below the table top. The work done in lifting it to the table top is equal to increase in its potential energy.
`W= ((Mg)/(n))(L)/(n)`
`:. W= (MgL)/(n^(2))`