question_answer

A lift of total mass M is raised by cables from rest through a height h. The greatest tension which the cables can safely bear is\[nMg\]. The maximum speed of lift during its journey if the ascent is to be made in shortest time is

A) \[\sqrt{2gh\left( \frac{n+1}{n} \right)}\]  

B)        \[\sqrt{2gh}\]                

C) \[\sqrt{2gh\left( \frac{n}{n+1} \right)}\]

D)        \[\sqrt{2gh\left( \frac{n-1}{n} \right)}\]

Correct Answer: D

Solution :

[d] Weight of lift = \[Mg\] Maximum tension = \[=nMg\] \[\therefore \]  Maximum acceleration                         \[=\frac{nMg-Mg}{M}=(n-1)g\] and maximum retardation = g Corresponding velocity-time graph for shortest time will be as follows: Here \[(n-1)\,g=\frac{{{v}_{m}}}{{{t}_{1}}}\]      or \[{{t}_{1}}=\frac{{{v}_{m}}}{(n-1)g}\] .....(1) And  \[g=\frac{{{v}_{m}}}{{{t}_{2}}}\]   or \[{{t}_{2}}=\frac{{{v}_{m}}}{g}\]                       .....(2) Area under v-t graph is total displacement h. Hence  \[h=\frac{1}{2}({{t}_{1}}+{{t}_{2}}){{v}_{m}}\]                             .....(3) From (1), (2) and (3) we get \[{{v}_{m}}=\sqrt{2gh\left( \frac{n-1}{n} \right)}.\]