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Uttarakhand PMT Uttarakhand PMT Solved Paper-2004

  • question_answer
    A conducting rod AC of length \[4l\] is rotated about \[_{B}^{a}\to \]point O in a uniform magnetic field B directed into the paper. \[AO=l\] and \[OC=3l\] then :

    A) \[VA-Vo=\frac{B\omega {{l}^{2}}}{2}\]

    B) \[Vo-Vc=\frac{9}{2}B\omega {{l}^{2}}\]

    C) \[Vo-Vc=4B\omega {{l}^{2}}\]

    D) \[Vo-Vc=\frac{9}{2}B\omega {{l}^{2}}\]

    Correct Answer: C

    Solution :

     \[{{V}_{O}}-{{V}_{A}}=\frac{B\omega l}{2}\times l\] \[=\frac{B\omega {{l}^{2}}}{2}\] \[{{V}_{O}}-{{V}_{C}}=\frac{B\omega 3l}{2}\times 3l\] \[=\frac{9B\omega {{l}^{2}}}{2}\] Hence, \[{{V}_{A}}-{{V}_{C}}=4B\omega {{l}^{2}}\] or        \[{{V}_{A}}>{{V}_{C}}\]


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