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A uniform chain of length L and mass M i...

A uniform chain of length `L` and mass `M` is tying on a smoth table and one third of its length is banging vertically down table the edge of the table if g is acceleration the to gravity , the work required to pull the hanging part on the table is

Text Solution

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Here, Weight of length `L` of the chain `=Mg`
Weight of length `(L)/(3)` of the chain which is hanging
`=(1)/(3)Mg`
The centre of gravity of the hanging part lies at its middle point, i.e.,, at a distance `=L//6` below the edge of the table.
Work required to pull the hanging part on the table,
`W=` force `xx` distance `=((1)/(3)Mg)xx(L)/(6)=(MgL)/(18)`
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Knowledge Check

  • A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part on to the table is

    A
    `MgL`
    B
    `MgL//3`
    C
    `MgL//9`
    D
    `MgL//18`
  • A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, work required to pull the hanging part on to the table is

    A
    `MgL`
    B
    `(MgL)/2`
    C
    `(MgL)/9`
    D
    `(MgL)/18`
  • A uniform chain of length l and mass m is placed on a smooth table with one-fourth of its length hanging over the edge. The work that has to be done to pull the whole chain back onto the table is :

    A
    `1/4 m gl`
    B
    `1/8 m gl`
    C
    `1/16 m gl`
    D
    `1/32 mgl`
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