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CMC Medical CMC-Medical VELLORE Solved Paper-2015

  • question_answer
    The electrostatics force of repulsion between two positively charged ions carrying equal charge is \[3.7\times {{10}^{-7}}N,\] when they are separated by a distance of 5 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]. What are the number of electrons are missing each ion?

    A)  2                            

    B)  4                            

    C)  5                            

    D)  10

    E)  7

    Correct Answer: A

    Solution :

                    From Coulambs law \[F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}}\] \[3.7\times {{10}^{-9}}=9\times {{10}^{9}}\times \frac{{{q}^{2}}}{{{(5\times {{10}^{-10}})}^{2}}}\] \[q=3.2\times {{10}^{-19}}c\] \[q=nc\] \[n=\frac{q}{c}=\frac{3.2\times {{10}^{-19}}}{1.6\times {{10}^{-10}}}=2\]


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