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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(m)/(sec^2)` and the maximum safe stress of the wire is `1xx10^7(N)/(m^2)`?A. 2 cmB. 1 cmC. 1.5 cmD. none of these |
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Answer» Correct Answer - A The tension T in the rope of the lift when it goes upward is given by `T=m(g+a)=314xx11N` Let r be the radius of the wire, Then maximum stress will be `(T)/(pir^2)` Hence `(T)/(pir^2)` `=1.1xx10^7` or `r^2=(T)/(pixx1.4xx10^8)=(314xx11)/(3.14xx(1.1xx10^7))=(1)/(10^4)` Now `r=(1)/(10^2)m=1cm` Diameter of the wire `=2r=2cm`. |
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