A heavy smooth chain of mass 'm' / length and total length 'L' hanging from smooth table edge. The length of chain on table is 'x''. The height of table top from ground is 'H'. If the motion starts from rest (t=0.0 sec) at what time the entire chain will descend on the ground.

Dear Bhaskar, We know the equation of motion as S = u + (1/2)a* t^2 Now the center of mass of the hanging part is at its midpiont and hence it is at height of H-[(L-x)/2] from the ground. And u(initial velocity)=0 a= g (acceleration due to gravity) , S=H-[(L-x)/2] Thus we would get time t= squareroot of{[(2*H)-(L-x)]/g} seconds Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best STUDENT NAME !!! Regards, Askiitians Experts Adapa Bharath