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A load of 1960 N is raised at the end of a steel wire. The minimum diameter of the wire so that stress in the wire does not exceed 100 N/mm2 is:

A. 4.0 mm

B. 4.5 mm

C. 5.0 mm

D. 5.5 mm

Answer: Option C


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 2 )

  1. Aziz Panna
    Aziz Panna :
    4 years ago

    Stress = (Load/Area)
    Therefore, Area= (Load/Stress) = 1960/100 = 19.6 mm^2
    Since, Area= pi × d^2 /4
    So, d = root of ( area × 4 / pi) = root of (19.6 ×4 / 3.14) = 5 mm

  2. Prakash Neupane
    Prakash Neupane :
    5 years ago

    Stress= load per unit area
    100 = 1960/(3.1416*d^2/4)
    Implies d= sqrt(1960*4/3.1416*100)
    = 5 mm

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