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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tang entail acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to \[8\times {{10}^{4}}\text{ J}\] by the end of the second revolution after the beginning of the motion?                                               

    A)  \[0.1\text{ }m/{{s}^{2}}\]                           

    B)  \[0.15\text{ }m/{{s}^{2}}\]

    C)   \[0.18\text{ }m/{{s}^{2}}\]                        

    D)   \[0.2\text{ }m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

                     \[\frac{1}{2}m{{v}^{2}}=E\Rightarrow \frac{1}{2}\left( \frac{10}{1000} \right){{v}^{2}}=8\times {{10}^{-4}}\]                 \[\Rightarrow \]    \[{{v}^{2}}=16\times {{10}^{-2}}\Rightarrow v=4\times {{10}^{-1}}=0.4\,\,m/s\]                 Now, \[{{v}^{2}}={{u}^{2}}+2{{a}_{t}}s\]                           \[(s=4\pi R)\] \[\Rightarrow \]               \[\frac{16}{100}={{0}^{2}}+2{{a}_{t}}\left( 4\times \frac{22}{7}\times \frac{6.4}{100} \right)\]                \[\Rightarrow \]               \[{{a}_{t}}=\frac{16}{100}\times \frac{7\times 100}{8\times 22\times 6.4}=0.1\,\,m/{{s}^{2}}\]


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