MPSC Rajyaseva 2017 GS Paper 1 — Question 52
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उदवाहकाची ओझे वाहून नेण्याची कमाल मर्यादा 1800 किलोग्राम (उदवाहकर + प्रवासी) आहे. हे उदवाहक्र उर्ध्व दिशेने 2 ms⁻¹ या एकसमान चालीने गतिमान आहे. गतिला विरोध करणारे घर्षण बल 4000 N आहे. तर मोटर यंत्राकळून उदवाहकाला किमान किती ताकत पुरवली गेली ते निश्चित करा. (g = 10 m/s²) An elevator carrying a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms⁻¹. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator. (Take g = 10 m/s²)
At constant speed the motor must balance gravity and friction: F = mg + f = 1800 × 10 + 4000 = 22,000 N. Power = F × v = 22,000 × 2 = 44,000 W. Since 1 hp = 746 W, that is 44,000 / 746 ≈ 59 hp.
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