Question

A 1000 kg lift is tied with metallic wires of maximum safe stress of $1.4\times {10}^{8}N/m^2$. If the maximum acceleration of the lift is $1.2c{m}^{-2}$, then the minimum diameter of the wire is

• A. $0.01m$
• B. $0.01cm$
• C. $0.001m$
• D. $0.02cm$

Answer 1

The correct option is A $0.01m$

Tension in wire$,$ $T=1000\left(9.8+1.2\right)=11000N$

$Maximumstress=\frac{T}{\pi {\left(s_{min}\right)}^2}$

$=11000\times \frac{7}{22\times {\left(s_{min}\right)}^2}$

$0r,{\left(s_{min}\right)}^2=\frac{{\left(50\right)}^2}{{\left({10}^4\right)}^2}$

$\therefore$ minimum radius $=50\times {10}^{-4}$

$\therefore$ minimum diameter $=2\times s_{min}$

$=2\times 50\times {10}^{-4}$

$=0.01m$

Hence,

option $\left(A\right)$ is correct answer.