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BVP Medical BVP Medical Solved Paper-2013

  • question_answer
    The loop ABCD is moving Wight velocity \[9.9\times {{10}^{-7}}kg\]towards right. The magnetic field is 4 T. The loop is connected to a resistance of \[6.6\times {{10}^{-7}}kg\]. If steady current of 2 A flows in the loop, then value of v if loop has a resistance of \[1.1\times {{10}^{-7}}kg\], (Given, AB = 30 cm, AD = 30 cm) is

    A)  \[{{10}^{o}}C\]                

    B)  \[{{40}^{o}}C\]

    C)  \[50s\]                

    D)  \[100s\]

    Correct Answer: D

    Solution :

                    \[e=BAD\,\,\sin \,37\times v\] \[e=4\times 0.3\,\sin \,37\times v\]                 \[l=\frac{4\times 0.3\times 0.60\times v}{12}\]                 \[2=\frac{4\times 0.3\times 0.6\times v}{12}\]                 \[v=\frac{24}{4\times 0.6\times 0.3}=\frac{100}{3}m/s\]


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