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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    A 30 gm bullet initially travelling at 120 m/s penetrates 12 cm into a wooden block. The average resistance exerted by the wooden block is [AFMC 1999; CPMT 2001]

    A)             2850N

    B)             2200 N

    C)             2000N

    D)             1800 N

    Correct Answer: D

    Solution :

                    \[F=\frac{m\,({{u}^{2}}-{{v}^{2}})}{2S}=\frac{30\times {{10}^{-3}}\times {{(120)}^{2}}}{2\times 12\times {{10}^{-2}}}=1800\,N\]


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