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JEE Main 7 Jan 2020 Shift 1 Solved Paper

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Question : 4

Total: 75

A 60 HP electric motor lifts an elevator havinga 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)

[7-Jan-2020 Shift 1]

1.7 ms‌–1
2.0 ms‌–1
1.9 ms‌–1
1.5 ms‌–1

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