A lift is tied with thick iron ropes having mass 'M'. The maximum acceleration of the lift is 'a'…

A lift is tied with thick iron ropes having mass 'M'. The maximum acceleration of the lift is 'a' $\mathrm{m} / \mathrm{s}^{2}$ and maximum safe stress is 's' $\mathrm{N} / \mathrm{m}^{2}$. The minimum diameter of the rope is ( $\mathrm{g}=$ acceleration due to gravity $)$
  1. $\left[\frac{2 \mathrm{M}(\mathrm{g}+\mathrm{a})}{\pi \mathrm{s}}\right]^{\frac{1}{2}}$
  2. $\left[\frac{2 \mathrm{M}(\mathrm{g}-\mathrm{a})}{\pi \mathrm{s}}\right]^{\frac{1}{2}}$
  3. $\left[\frac{4 \mathrm{M}(\mathrm{g}+\mathrm{a})}{\pi \mathrm{s}}\right]%{\frac{1}{2}}$
  4. $\left[\frac{4 \mathrm{M}(\mathrm{g}-\mathrm{a})}{\pi \mathrm{s}}\right]^\frac{1}{2}$

Solution

The maximum stress produced in a rope is given by \(\sigma_{\max }=\frac{\text { Force }}{\text { Area }}=\frac{M g}{\pi r^2}\) As the lift is accelerating with acceleration \(a\), then $\begin{aligned} & \sigma_{\max}=\frac{M(g \pm a)}{\pi r^{2}} \\ & \Rightarrow \quad r^{2}=\frac{M(g \pm a)}{\pi S} \quad\left[\text{Given, } \sigma_{\max}=S\right] \\ & \frac{d^{2}}{4}=\frac{M(g \pm a)}{\pi S} \quad\left[\because r=\frac{d}{2}\right] \\ & \Rightarrow \quad d=\sqrt{\frac{4 M(g \pm a)}{\pi S}} \end{aligned}$ As acceleration is maximum i.e., \(g^{\prime}=g+a\), so \(d=\sqrt{\frac{4 M(g+a)}{\pi S}}\)

Asked in: MHT CET 2020 (15 Oct Shift 2)

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