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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through \[{{60}^{o}}\]. The torque needed to maintain the needle in this position will be     AIEEE  Solved  Paper-2003

    A) \[\sqrt{3}\] W

    B)                       W           

    C)       \[(\sqrt{3}/2)\]W                            

    D) 2 W

    Correct Answer: A

    Solution :

    As the work \[W=MB\,(1-\cos \theta )\] \[\Rightarrow \]   \[W=MB\,(1-\cos \,{{60}^{o}})\]               \[(\because \theta ={{60}^{o}})\] \[\Rightarrow \]   \[W=\frac{MB}{2}\] \[\therefore \]      MB = 2 W Torque, \[\tau =MB\sin {{60}^{o}}\]                     \[=\frac{MB\sqrt{3}}{2}=\frac{2W\sqrt{3}}{2}=W\sqrt{3}\]


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