Question

. A uniform chain of length L and mass M is lying on
a smooth table and 2/3 of its length is hanging down
over the edge of the table. If g is the acceleration
due to gravity, the work done to pull the hanging
part on the table is :-
(1) Mgl
(2) MgL/3
(3) Mgl
(4) 2MgL/9​

Answer 1

Answer:Mgl/18

Explanation: work done is force*displacement

Also center of mass is at Mg/n if nth path was hanging and length is also l/2n

So formula gets modified into

Mgl/2n(squar)

n is the fraction of hanging part

Here 1-2/3 it's 1/3th part hanging

So mgl/2*({3)square}

It's mgl/18

It was an aakash all time favourite question

Hope you got this!

Answer 2

apply direct formula mgl/2n^2

m*10*l/2*(3)^2

10lm/18=5lm/9