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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, the work required to pull the hanging part on to the table is
  • a)
    MgL 
  • b)
    MgL/3 
  • c)
    MgL/9
  • d)
    MgL/18
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A uniform chain of length L and mass M is lying on a smooth table and ...
 The hanging part of the chain which is to be pulled up can be considered as a point mass situated at the centre of the hanging part. The equivalent diagram is drawn.
Note : The work done in bringing the mass up will be equal to the change in potential energy of the mass.
W = Change in potential energy
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Most Upvoted Answer
A uniform chain of length L and mass M is lying on a smooth table and ...
To understand why the correct answer is option 'D', let's break down the problem step by step.

1. Finding the mass of the hanging part:
- We are given that one third of the chain's length is hanging vertically down over the edge of the table.
- Since the chain is uniform, we can assume that the mass is distributed uniformly along its length.
- Therefore, the mass of the hanging part is (1/3) * M, where M is the total mass of the chain.

2. Finding the work required to lift the hanging part onto the table:
- When the hanging part is lifted onto the table, the height through which it is lifted is equal to the length of the hanging part.
- Let's denote the length of the hanging part as L/3, where L is the total length of the chain.
- The work done in lifting the hanging part is equal to the change in potential energy.
- The change in potential energy is given by the formula: ΔPE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height.
- Substituting the values we have, the work done is (1/3) * M * g * (L/3).

3. Simplifying the expression:
- We can simplify the expression further by canceling out common factors.
- (1/3) * M * g * (L/3) = M * g * (L/9).

Therefore, the work required to pull the hanging part onto the table is M * g * (L/9), which corresponds to option 'D'.
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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, the work required to pull the hanging part on to the table isa)MgLb)MgL/3c)MgL/9d)MgL/18Correct answer is option 'D'. Can you explain this answer?
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