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Manipal Medical Manipal Medical Solved Paper-2004

  • question_answer
    A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 100 m/s with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against the gravitational attraction? (Take\[g=10\text{ }m/{{s}^{2}}\]).

    A)  50 kg/s     

    B)  100 kg/s

    C)  200 kg/s    

    D)  400 kg/s

    Correct Answer: B

    Solution :

     Given: Rockets mass\[m=100\text{ }kg\] The velocity of exhaust gases with respect to rocket\[=10\text{ }m/s\] \[g=10\text{ }m/{{s}^{2}}\] The minimum force on the rocket to lift it \[{{F}_{\min }}=mg=1000\times 10=10000\text{ }N\] Hence, minimum rate of burning of fuel is given by \[\frac{dm}{dt}=\frac{{{f}_{\min }}}{\upsilon }=\frac{10000}{100}\] \[=100kg/s\]


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