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BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    A bullet is fired from a gun. The force on the bullet is given by\[F=600-2\times {{10}^{5}}t,\]where F is in newton and t is in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

    A)  9Ns                                      

    B)  Zero

    C)  0.9 Ns                  

    D)  1.8 Ns

    Correct Answer: C

    Solution :

                      \[F=600-2\times {{10}^{5}}t\] At \[t=0,\]           \[F=600N\] As\[F=0,\]on leaving the barrel, \[\therefore \]  \[0=600-2\times {{10}^{5}}t\] This is the time spent by the bullet in the barrel. Average force\[=\frac{600+0}{2}=300\text{ }N\] \[\therefore \] Average impulse imparted \[=F\times t\] \[=300\times 3\times {{10}^{-3}}=0.9Ns\]


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