A lift is tied with thick iron and its mass is 314 kg. What should be the minimum diameter of wire if the maximum acceleration of lift is 1.2 $\mathrm{m}/{\sec }^2$ and the maximum safe stress of the wire is 1 x ${10}^7 \mathrm{N}/{\mathrm{m}}^2$?

The tension, T in the rope of the lift when it goes upward is given by

$\mathrm{T} = \mathrm{m}(\mathrm{g}+\mathrm{a}) = 314\times 11\mathrm{N}$

Let r be the radius of the wire, then maximum stress will be $\frac{\mathrm{T}}{{\mathrm{\pi r}}^2}$

Hence, $\frac{\mathrm{T}}{{\mathrm{\pi r}}^2} = 1.1\times {10}^7$

or  ${\mathrm{r}}^2 = \frac{\mathrm{T}}{\mathrm{\pi }\times 1.4\times {10}^8} = \frac{314\times 11}{3.14\times (1.1\times {10}^7)} = \frac{1}{{10}^4}$

Now  $\mathrm{r} = \frac{1}{{10}^2}\mathrm{m} = 1 \mathrm{cm}$

$\therefore$ Diameter of the wire = 2r = 2cm.