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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

  • question_answer
        Four resistances of\[10\,\Omega ,60\,\Omega ,100\,\Omega \]and\[200\,\Omega \]. respectively taken in order are used to form a Wheatstones bridge. A 15 V battery is connected to the ends of a\[200\,\Omega \] resistance. The current through it will be

    A)  \[7.5\times {{10}^{-5}}A\]          

    B)  \[7.5\times {{10}^{-4}}A\]

    C)  \[7.5\times {{10}^{-3}}A\]          

    D)  \[7.5\times {{10}^{-2}}A\]

    Correct Answer: D

    Solution :

                    Here, the resistances\[10\,\Omega ,60\,\Omega \] and\[100\,\Omega ,\]are in series and they together are in parallel to \[200\,\Omega ,\]resistance. When a potential difference of 15 V is applied across\[200\,\Omega ,\]then current through it is                 \[I=\frac{15}{200}=7.5\times {{10}^{-2}}A\] Magnetic field at the centre of circular loop.


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