A CHAIN OF MAAS M AND LENGHT L IS KEPT VERTICAL SO THAT ITS LOWEST POINT JUST TOUCHES THE GROUND IF THE UPPER END OF THE CHAIN IS RELEASED FIND THE NORMAL FORCE EXERTED BY THE GROUND ON THE CHAIN WHEN THE HIGHEST LINK OF THE CHAIN FALLS DOWN BY A DISTANCE x

plz give easy expressions with full explanation calculus is weak xi standard i approve

Hi Raj

The Question says that initially the chain just touches the ground which means that initially N(normal force)=0

Now mass of length L of the chain is M

so mass of 1 unit lenght of the chain will be M/L

therefore mass of a small elemental part of the chain of lenght dx= M/L dx

now when the top most part of the chain falls down by a distance x.....the length of chain on the ground is also x

now Normal force = mass x g

so Normal force on the elemental part of lenght dx = M/L g dx

integrating this expression from 0 to x to get normal force on x lenght of the chain will be ∫dN=∫M/L g dx

N=M/L g ∫dx ( M, L and g are constants and so can be taken out of the integral)

N=Mgx/L

Hope you liked my answer!!

All the Best!!