BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    A particle is moving in a circle of radius R with constant speed v. If radius is doubled, then its centripetal force to keep the same speed gets:

    A)  twice as great as before

    B)         half

    C)         one-fourth

    D)         remains constant

    Correct Answer: B

    Solution :

    Centripetal force is given by \[F=\frac{m{{v}^{2}}}{R}\] where m is mass of particle, \[v\]is speed, and R is radius of circular path. \[\Rightarrow \]               \[F\propto \frac{1}{R}\] or            \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] Given,   \[{{R}_{2}}=2{{R}_{1}}\] \[\therefore \]  \[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{{{R}_{1}}}{2{{R}_{1}}}=\frac{1}{2}\] or            \[{{F}_{2}}=\frac{{{F}_{1}}}{2}\] Therefore, centripetal force will become half.


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