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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Topic Test - Newton's Laws Of Motion

  • question_answer
    If M is mass of rocket, r is rate of ejection of gases with respect to rocket, then acceleration of the rocket \[\frac{dv}{dt}\] is equal to  

    A) \[\frac{ru}{(m-rt)}\]

    B) \[\frac{(m-rt)}{ru}\]

    C) \[\frac{ru}{(m+rt)}\]

    D) \[\frac{ru}{m}\]

    Correct Answer: A

    Solution :

    [a] Here, initial mass of rocket = M. \[\frac{dm}{dt}=\,r\] Relative velocity of gases w.r.t. rocket = v then acceleration of the rocket, \[a\,=\,\frac{F}{m}=\,\frac{u\,(dm/dt)}{\left( M-\frac{dm}{dt}\times \,t \right)}\,=\frac{ur}{M-rt}\]


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