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JEE Main 2020 (7 Jan, Shift 1) · Power

Power question from JEE Main 2020 (7 Jan, Shift 1)

Q32JEE Main 2020 · 7 JAN · SHIFT 1Medium

Power

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2)

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Answer: (d)

  1. P=60×746=44760 WP = 60 \times 746 = 44760\text{ W}.
  2. Force needed F=mg+f=20000+4000=24000 NF = mg + f = 20000 + 4000 = 24000\text{ N}.
  3. v=PF=4476024000≈1.87 m/sv = \dfrac{P}{F} = \dfrac{44760}{24000} \approx 1.87\text{ m/s}
  4. So the answer is (d) 1.9 m s−11.9\text{ m s}^{-1}.
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