Punjab Medical Punjab - MET Solved Paper-2007

  • question_answer
    A voltmeter having a resistance of \[998\Omega \] is connected to a cell of emf \[2V\] and internal resistance\[2\Omega \]. The error in the measurement of emf will be

    A) \[4\times {{10}^{-1}}V\]                               

    B) \[2\times {{10}^{-3}}V\]

    C) \[4\times {{10}^{-3}}V\]                               

    D) \[2\times {{10}^{-1}}V\]

    Correct Answer: C

    Solution :

    Current taken from the cell                 \[i=\frac{E}{R+r}\] where\[E=\]emf of cell \[R=\]external resistance \[r=\]internal resistance \[\therefore \]  \[i=\frac{2}{998+2}=\frac{1}{500}A\] Error in value of emf = Actual value of emf - measured emf Measured emf \[V=E-ir\]                 \[=2-\frac{1}{500}\times 2\]                 \[=2-\frac{1}{250}\] \[\therefore \]  Error\[=2-\left( 2-\frac{1}{250} \right)=\frac{1}{250}\]                 \[=4\times {{10}^{-3}}V\]


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