A lift of total mass `M` is raised by cable from rest through a height `h` . The greatest tension which the cables car safely bear is `n` Mg. The maximum sped of lift during its journey if the ascent is to made in shortest time is

Options

A. `sqrt(2gh((n+1)/(n))`
B. `sqrt(2gh)`
C. `sqrt(2gh((n)/(n+1)))`
D. `sqrt(2gh((n-1)/(n)))`

Answer

D

Explanation

Weight of lift=mg
Maximum tension `=nMg`
:. Maximum acceleration `=(nMg-Mg)/(M)`
`=(n-1)g`
and maximum retardation=g
corresponding velocity-time graph for shortest time will be as follows
Here `(n-1)g=(v_(m))/(t_(1))` or `t_(1)=(v_(m))/((n-1)g)` (1)
and `g=(v_(m))/(t_(2))` or `t_(2)=(v_(m))/(g)` ...(2)
Area under `v-t` graph is total displacement `h` .
Hence `h=(1)/(2)(t_(1)+t_(2))v_(m)` ...(3)
From (1), (2) and (3) we get,
`v_(m)=sqrt((2gh(n-1)/(n))` .