A lift of mass 'm' is connected to a rope which is moving upward with maximum acceleration 'a'. For maximum safe stress, the elastic limit of the rope is 'T'. The minimum diameter of the rope is
(g = gravitational acceleration)

Options

A. `[(2m(g+a))/(piT)]^(1//2)`
B. `[(4m(g+a))/(piT)]^(1/2)`
C. `[(m(g+a))/(piT)]^(1/2)`
D. `[(m(g+a))/(2piT)]^(1/2)`

Answer

B

Explanation

`((omega)/(4))^(2)=omega^(2)-2alphan(2pi)`
`therefore 2alphan(2pi)=omega^(2)-(omega^(2))/(16)`
`therefore 2pin=(15)/(16)((omega^(2))/(2alpha))`
`0=omega^(2)-2alphan'`
`therefore 2pin'=(omega^(2))/(2alpha)" "therefore n'=(16)/(15)n`
`T=(4m(g+a))/(pid^(2))`
`therefore d^(2)=(4m(g+a))/(piT)`
`d=[(4m(g+a))/(piT)]^(1/2)`