JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Power

  • question_answer
    A quarter horse power motor runs at a speed of 600 r.p.m. Assuming 40% efficiency the work done by the motor in one rotation will be                     [Kerala PET 2002]

    A)             7.46 J 

    B)             7400 J

    C)             7.46 ergs

    D)               74.6 J

    Correct Answer: A

    Solution :

                Motor makes 600 revolution per minute             \ n = \[600\frac{\text{revolution}}{\text{minute}}=10\frac{rev}{\sec }\]             \ Time required for one revolution \[=\frac{1}{10}\]sec             Energy required for one revolution = power × time                                     =\[\frac{1}{4}\times 746\times \frac{1}{10}=\frac{746}{40}J\]             But work done = 40% of input                                     \[=40%\times \frac{746}{40}=\frac{40}{100}\times \frac{746}{40}=7.46\,J\]


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