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

  • question_answer
    Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

    A) \[\frac{\sqrt{3}}{4}N\]          

    B)  \[\sqrt{3}\,N\]

    C)  0.5 N             

    D)  1.5N

    Correct Answer: C

    Solution :

     Taking x-components, the total should be zero. \[\frac{Q}{4\pi {{R}^{2}}}=\frac{q}{4\pi {{r}^{2}}}=\sigma \] \[\therefore \] \[V=\frac{1}{{{\varepsilon }_{0}}}\,[\sigma R+\sigma r]=\frac{\sigma }{{{\varepsilon }_{0}}}[R+r]\] This is the required force in x-direction.

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