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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    A magnetic needle suspended parallel to a magnetic field requires \[\sqrt{\text{3}}\,\text{J}\] of work to turn it through\[60{}^\circ \]. The torque needed to maintain the needle in this position will be                             

    A) \[2\sqrt{\text{3}}\,\text{J}\]     

    B) \[3\,\text{J}\]                   

    C) \[\sqrt{\text{3}}\,\text{J}\]                        

    D) \[\frac{3}{2}\,\text{J}\]

    Correct Answer: B

    Solution :

    In this case, work done \[W=MB(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\] \[=MB(\cos {{0}^{o}}-\cos {{60}^{o}})\] \[=MB\left( 1-\frac{1}{2} \right)=\frac{MB}{2}\] \[\text{MB}\,\text{=}\,\text{2}\sqrt{\text{3}}\,\text{J}\]\[(\because \text{given}\,\text{W=}\sqrt{\text{3}}\,\text{J})\] \[\text{ }\!\!\tau\!\!\text{ }\,\text{=}\,\text{MB}\,\text{sin}\,\text{6}{{\text{0}}^{\text{o}}}\text{=}\,\text{(2}\sqrt{\text{3}}\text{)}\left( \frac{\sqrt{\text{3}}}{\text{2}} \right)\text{J}\,\text{=}\,\text{3}\,\text{J}\]


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