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

The maximum tesion in the rope
`= m(g +a)`
Stress in the rope, `T = (m(g+a))/(pir^(2))`
`:. T = (m(g+a))/(pir^(2)) = (m(g+a))/(pi(d/2)^(2))`
`rArr T = (4m(g+a))/(pid^(2))`
`rArr d^(2) = (4m(g+a))/(piT)`
`:. d = [(4m(g+a))/(piT)]^(1//2)`