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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    A particle moves along a circle of radius \[\left( \frac{20}{\pi } \right)\,m\]with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun the tangential acceleration will be

    A) \[40\pi \text{ }m/{{s}^{2}}\]         

    B)       \[~40\text{ }m/{{s}^{2}}\]           

    C)        \[~160\pi \,m/{{s}^{2}}\]       

    D)         \[~640\pi \,m/{{s}^{2}}\]

    Correct Answer: B

    Solution :

    We have, \[{{v}^{2}}={{u}^{2}}+2as\] \[\Rightarrow \]               \[{{(80)}^{2}}=0+2(a)\left( 4\pi \times \frac{20}{\pi } \right)\]  [as two revolutions are completed] \[\Rightarrow \]               \[a=40\,m/{{s}^{2}}\]


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