JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A charge \[Q\] is divided into two parts q and q' and separated by a distance R. The force of repulsion between them will be maximum, when

    A)  \[q=\frac{Q}{4}\]                

    B)  \[q=\frac{Q}{2}\]

    C)  \[q=Q\]                     

    D)  None of these

    Correct Answer: B

    Solution :

     As total charge \[Eq\] \[{{T}_{1}}=15a\]    \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{15a}{3a}=\frac{5}{1}\] Force between the charges q and q' \[\Rightarrow \] For maximum value of F \[{{T}_{1}}:{{T}_{2}}=5:1\] or \[T=M\,\left( g-\frac{g}{4} \right)=\frac{3Mg}{4}\] \[W=\mathbf{T}\cdot \mathbf{d}\Rightarrow \,W=Td\]       \[\Rightarrow \]or         \[W=-Td=-\frac{3Mgd}{4}\]

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