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Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    A thin bar magnet of length 2L is bent at the mid-point so that the angle between them is\[60{}^\circ \]. The new length of the magnet is:

    A)  \[\sqrt{2}L\]          

    B)  \[\sqrt{3}L\]

    C)  \[2L\]   

    D)  L

    Correct Answer: D

    Solution :

     On bending the magnet, the length of the magnet \[AC=AB+BC\] \[=L\sin \left( \frac{\theta }{2} \right)+L\sin \left( \frac{\theta }{2} \right)\] \[=L\sin \left( \frac{\theta }{2} \right)\] \[=2L\sin 30{}^\circ \] \[=2L\times \frac{1}{2}=L\]


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