In the above question, the maximum power delivered by the agent in pul

1.	In the above question, the maximum power delivered by the agent in pulling up the rope is
Answer» `lambda//gv` (B) `lambda//gv+(v^3lambda)/(2)` (c) `lambda//gv+v^3lambda` (d) `(llgv)/(2)+(lambdav^3)/(2)` Solution :Instantaneous force `=mg+(vdm//dt)`, where `mg` is the force needed for suppording the weight of already hanging section of the rope and `vdm//dt` is the force needed to SUPPLY momentum to the portion of the rope which is to be pulled up. `(mg)_(max)=(lambdal)g, (DM)/(dt)=lambdav` Hence, `F_(max)=lambdalg+lambdav^2` and `P_(max)=lambdalgv+lambdav^3`

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