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JEE Main 20207 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2020 · 7 January · Morning Shift · Physics · Laws of Motion

A 60HP electric motor lifts an elevator having a maximum total load...

A 60HP60HP electric motor lifts an elevator having a maximum total load capacity of 2000kg.2000kg . If the frictional force on the elevator is 4000N,4000N, the speed of the elevator at full load is close to : (1HP=746W,g=10ms−2)\left(1HP = 746W, g = 10m{s}^{- 2}\right)

Options

  1. A1.7ms−11.7m{s}^{- 1}
  2. B1.9ms−11.9m{s}^{- 1}
  3. C1.5ms−11.5m{s}^{- 1}
  4. D2.0ms−12.0m{s}^{- 1}

Correct answer

B. 1.9ms−11.9m{s}^{- 1}

Step-by-step solution

4000×V+mg×V=P4000 \times V + mg \times V = P
60×7464000+20000=V\dfrac{60 \times 746}{4000 + 20000} = V
V=V =1.9ms−11.9m{s}^{- 1}
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