Discussion Forum



1)

A lift of mass 'm' is connected to a rope that 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)


A) $\left[\frac{2m(g+a)}{\pi T}\right]^{1/2}$

B) $\left[\frac{4m(g+a)}{\pi T}\right]^{1/2}$

C) $\left[\frac{m(g+a)}{\pi T}\right]^{1/2}$

D) $\left[\frac{m(g+a)}{ 2\pi T}\right]^{1/2}$

Answer:

Option B

Explanation:

 The maximum tension in the rope =m(g+a)

Stress in the rope ,  $T=\frac{m(g+a)}{\pi r^{2}}$

$\therefore$    $T=\frac{m(g+a)}{\pi r^{2}}=\frac{m(g+a)}{\pi (\frac{d}{2})^{2}}$

 $\Rightarrow$    $ T=\frac{4m(g+a)}{\pi d^{2}}$

$\Rightarrow$    $ d^{2}=\frac{4m(g+a)}{\pi T}$

$\therefore$     d=$\left[\frac{4m(g+a)}{\pi T}\right]^{1/2}$