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VMMC VMMC Medical Solved Paper-2010

  • question_answer
    A 10 m long wire of resistance 20\[\Omega \]  is connected in series with a battery of emf 3V and a resistance of 10\[\Omega \]. The potential gradient along the wire in V/m is

    A)  0.02      

    B)                         0.1                        

    C)         0.2                        

    D)         1.2

    Correct Answer: C

    Solution :

    As resistances are in series \[{{R}_{total}}={{R}_{1}}+{{R}_{2}}\] \[=20+10=30\Omega \] \[i=\frac{4}{{{R}_{\text{total}}}}=\frac{3}{30}=\frac{1}{10}\text{A}\]                 So           \[{{V}_{\text{wire}}}=\,\,i{{R}_{\text{wire}}}\]                 \[\Rightarrow \]               \[{{V}_{\text{wire}}}=\frac{1}{10}\times 20=2V\] Hence, potential gradient is \[\frac{{{V}_{\text{wire}}}}{l}=\frac{2}{10}=0.2\,\,\text{V/m}\]


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