$\begin{array}{1 1}(A)\;.01m\\(B)\;.005m\\(C)\;.5m\\(D)\;\text{None of above}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

If the maximum acceleration of lift is $2m/s^2$ and then maximum force is $m(g+a)$

$\Rightarrow 1500\times 12$

$\Rightarrow 18000N$

Force on 1wire=$\large\frac{18000}{4}$N

$\Rightarrow 4500N$

Maximum stress =$\large\frac{4500}{\pi r^2}$

This should be less than given safe stress

$\large\frac{4500}{\pi r^2}$$ < 2\times 10^8$

$716.2\times 10^{-8} < r^2$

$26.76\times 10^{-4} < r$

$.00535 < D$

Hence (B) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...