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 \[{{v}_{m}}=\sqrt{2gh\left( \frac{n-1}{n} \right)}.\]You need to login to perform this action.
You will be redirected in
3 sec