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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    A galvanometer of resistance \[100\Omega \]has 50 divisions on its scale and has sensitivity of \[20\mu A/\]division. It is to be converted to a voltmeter with three ranges, of 0-2 V, 0-10 V and 0-20 V. The appropriate circuit to do so is :                             [JEE Main Held on 12-4-2019 Morning]

    A)

    B)

    C)

    D)

    Correct Answer: D

    Solution :

    \[20\times 50\times {{10}^{6}}={{10}^{3}}Amp.\] \[{{V}_{1}}=\frac{2}{{{10}^{-3}}}=100+{{R}_{1}}\]                       \[1900={{R}_{1}}\] \[{{V}_{2}}=\frac{10}{{{10}^{-3}}}=(2000+{{R}_{2}})\]    \[{{R}_{2}}=8000\] \[{{V}_{3}}=\frac{20}{{{10}^{-3}}}=10\times {{10}^{3}}+{{R}_{3}}\]\[10\times {{10}^{3}}={{R}_{3}}\]


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