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Manipal Medical Manipal Medical Solved Paper-2009

  • question_answer
    A quarter horse power motor runs at a speed of 600 rpm. Assuming 40% efficiency the work done by the motor in one rotation will be

    A)  7.46 J          

    B)  7400 J

    C)  7.46 erg        

    D)  74.6 J

    Correct Answer: A

    Solution :

    \[P\times 40%=\frac{W}{t}\] \[\Rightarrow \] \[\frac{746}{4}\times \frac{W}{t}=\left( \frac{W}{2\pi \times \frac{600\times 2\pi }{60}} \right)\] \[\therefore \] \[W=7.46\,J\]


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