A chain AB of length l is lying in a smooth horizontal tube so that the fraction 'h' of its length hangs freely and just touches the surface of the table with its end B. At a certain moment the end A of the chain is set free. The velocity of end A of the chain when it just slips out of tube is :
` (##MOD_RPA_OBJ_PHY_C04_A_E01_089_Q01.png" width="80%">

Options

A. `hsqrt((2g)/(lh))`
B. `sqrt(2ghlog(l/(h))`
C. `sqrt(2gllog(l/(h))`
D. None of these

Answer

b

Explanation

Here let M be the total mass of the chain `(M)/(l)`and themass per unit length. Then accelerating force is weight of the hanging part. For a certain length x of the chain, the equation of motion is :
`(M)/(l)xxhxxg=(M)/(l)xxXxx(dupsilon)/(dt)`
or`hg.dx=x(dupsilon)/(dt).dx`
or`int_(h)^(l)hg(dx)/(x)=int_(0)^(v) nu dnu`
`[log_(e)x]_(h)^(l)=(upsilon^(2))/(2)`or`upsilon=(2hg log_(e) (l)/(h))^(1//2)`
(b) is the choice.