BVP Medical BVP Medical Solved Paper-2004

  • question_answer
    A bomber plane is moving horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which the bomb strikes the ground is, (g = 10 \[2.4\times {{10}^{-9}}N\]) :

    A) \[1.5\times {{10}^{-3}}N\]                           

    B) \[1.5\times {{10}^{-3}}N\]

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

    D) \[m/{{s}^{2}}\]

    Correct Answer: C

    Solution :

    Speed of plane \[\Omega \] Time taken by the bomb to strike the ground \[I=100\sin 200\pi t\]      The time taken is given by                 \[t=\frac{\sqrt{2h}}{g}\] or \[\frac{\sqrt{2h}}{g}=10\] or            \[\frac{1}{100}s\] or            \[\frac{1}{200}s\] or            \[L=\frac{0.4}{\pi }\]       Now vertical velocity is given by \[R=30\Omega \] Hence, \[\Omega \] or \[\Omega \]


You need to login to perform this action.
You will be redirected in 3 sec spinner