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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through\[{{60}^{o}}\]. The torque required to keep the needle in this position will be

    A)  \[2W\]                

    B)         \[W\]                  

    C)  \[\frac{W}{\sqrt{2}}\]                  

    D)         \[\frac{W}{\sqrt{3}}\]

    E)  \[\sqrt{3}W\]

    Correct Answer: E

    Solution :

    \[W=mB\text{ }cos\theta =mB\text{ }cos\text{ }{{60}^{o}}\] \[=mB\times \frac{1}{2}\] \[\tau =mB\sin \theta \] \[=mB\sin {{60}^{o}}\] \[=\sqrt{3}W\]


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