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A cylindrical steel wire of `3 m` length is to stretch no more than `0.2 cm` When a tensile force of `400 N` is applied to each end of the wire ? What minimum diameter is required for the wire ?? `Y_(steel) = 2.1xx10^(11) N//m^2` |
Answer» Correct Answer - A::C `Deltal = (Fl)/(AY) = (Fl)/((pid^(2)//4)Y)` `:. d = sqrt((4 Fl)/(pi(Deltal)Y))` ` = sqrt((4xx400xx3)/(3.14xx0.2xx10^(-2)xx2.1xx10^(11)))` `= 1.91 xx10^(-3) m = 1.91 mm` |
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