JEE PYQ: Laws Of Motion Question 19
Question 19 - 2020 | 07 Jan Shift 1] (MCQ)
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, $g = 10,\text{ms}^{-2}$)
(1) 1.7 ms$^{-1}$
(2) 1.9 ms$^{-1}$ [Correct]
(3) 1.5 ms$^{-1}$
(4) 2.0 ms$^{-1}$
Show Answer
Answer: (2)
Solution
$F = 24000$ N. $P = Fv$. $v = \frac{44760}{24000} \approx 1.9$ m/s.
BETA
