JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the \[{{\text{n}}^{\text{th}}}\] power of R. If the period of rotation of the particle is T, then: [JEE Main Online 08-04-2018]

    A)  \[\text{T}\propto {{\text{R}}^{(n+1)/2}}\]                       

    B)  \[\text{T}\propto {{\text{R}}^{n/2}}\]

    C)  \[\text{T}\propto {{\text{R}}^{3/2}}\] for any \[n\]

    D)  \[T\propto {{R}^{\frac{n}{2}+1}}\]

    Correct Answer: A

    Solution :

    For circular motion \[\frac{m{{v}^{2}}}{R}=\frac{c}{{{R}^{n}}}\] \[\therefore v=\sqrt{\frac{c}{m{{R}^{n-1}}}}\] \[\therefore Time\,\,period\,\,T=\left( \frac{2\pi R}{v} \right)\] \[T\propto {{R}^{\left( \frac{n+1}{2} \right)}}\]


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