2. Solved Papers
  3. JIPMER

JIPMER Jipmer Medical Solved Paper-2015

  • question_answer
    The work done in turning a magnet of magnetic moment M by an angle of \[90{}^\circ \] from the meridian, is n times the corresponding work done to turn it through an angle of\[60{}^\circ \]. The value of n is given by:

    A)  2                                            

    B)  1

    C)  1/2                                       

    D)  1/4

    Correct Answer: A

    Solution :

    Given: Angle of magnet \[(\theta )=90{}^\circ \]and\[60{}^\circ \]. We know that work done in turning the magnet through\[90{}^\circ \]. \[({{W}_{1}})=MB\,(\cos 0{}^\circ -\cos 90{}^\circ )\]      \[=MB\,(1-0)=MB.\]  Similarly\[{{W}_{2}}=MB\,(\cos 0{}^\circ -\cos 60{}^\circ )\]                  \[=MB\left( 1-\frac{1}{2} \right)=\frac{MB}{2}.\] Therefore, \[{{W}_{1}}=2\,{{W}_{2}}\]or\[n=2\,.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner