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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%"> |
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Answer» `hsqrt((2g)/(lh))` `(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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