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Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    A short bar magnet is placed with its south pole towards geographical north. The neutral points are situated at a distance of 20 cm from the   centre   of   the   magnet.    If \[f(x)=3\sin x-4{{\sin }^{3}}x\]then the magnetic moment of the magnet is

    A)  9000ab - amp x cm2       

    B) 900ab - amp x cm2

    C) 1200ab - amp x cm2        

    D) 225ab - amp x cm2

    Correct Answer: C

    Solution :

    At neutral point magnetic field due to magnet = Horizontal component of the earth's magnetic  field. \[\frac{37}{9}M{{R}^{2}}\]\[\frac{40}{9}M{{R}^{2}}\] \[~50m/{{s}^{2}}\]\[50.5m/{{s}^{2}}\] \[51m/{{s}^{2}}\]\[42m/{{s}^{2}}\]


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