Question

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)

विकल्प

• `[(2"m"("g"+"a"))/(pi"T")]^(1/2)`

• `[(4"m"("g"+"a"))/(pi"T")]^(1/2)`

• `[("m"("g"+"a"))/(pi"T")]^(1/2)`

• `[("m"("g"+"a"))/(2pi"T")]^(1/2)`

Answer 1

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 `underlinebb([(4"m"("g"+"a"))/(pi"T")]^(1/2))`.

(g = gravitational acceleration)

Explanation:

Maximum tension in the rope = m (g + a)

Stress rn the rope, T = `("m"("g"+"a"))/(pi"r"^2)`

∴ T = `("m"("g"+"a"))/(pi"r"^2)=("m"("g"+"a"))/(pi("d"/2)^2)`

or, T = `(4"m"("g"+"a"))/(pi"d"^2)` 

⇒ d² = `(4"m"("g"+"a"))/(pi"T")`  

∴ d = `[(4"m"("g"+"a"))/(pi"T")]^(1/2)`