1.

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%">

Answer»

`hsqrt((2g)/(lh))`
`sqrt(2ghlog(L/(h))`
`sqrt(2gllog(l/(h))`
None of these

Solution :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.


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