In the above question, the average power delivered by gravity is

Options

A. `lambda//gv`
B. (b) `lambda//gv+(v^3lambda)/(2)`
C. (c) `lambda//gv+v^3lambda`
D. (d) `(llgv)/(2)+(lambdav^3)/(2)`

Answer

C

Explanation

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`