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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 the acceleration due to gravity, work required to pull the hanging part into the table is: MgL MgL/3 MgL/9 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 the acceleration due to gravity, work required to pull the hanging part into the table is:
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  1. MgL
  2. MgL/3
  3. MgL/9
  4. MgL/18

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Grade:11

2 Answers

Abhishek Anand
40 Points
6 years ago
As L/3 length is variable. So we assume it `x`.Then mass of hanging chain be M(x)/L. This is also a variable force by which we act on chain for take whole chain on table. As this force is variable so work done is W=(int)Fdx, where (int=integration). Then equation be W=(int) {Mg(x)/L}dx where integration is from 0 to L/3 ,After integration, W=Mg(x^2/2)/L ,Then put limit 0 to L/3 , we get MgL/18 , which is our answer ....Option D is correct.
Abhishek Anand
40 Points
6 years ago
As L/3 length is variable. So we assume it `x`.Then mass of hanging chain be M(x)/L. Force is Mg(x)/L which is also a variable force by which we act on chain for take whole chain on table. As this force is variable so work done is W=(int)Fdx, where (int=integration). Then equation be W=(int) {Mg(x)/L}dx where integration is from 0 to L/3 ,After integration, W=Mg(x^2/2)/L ,Then put limit 0 to L/3 , we get MgL/18 , which is our answer ....Option D is correct.

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