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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2020
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  • MCQ

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

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Answer & Solution

Correct Answer

(C) \(1.9\)

Step-by-step Solution

Detailed explanation

Let elevator is moving upward with constant speed V.
Tension in cable

\(\mathrm{T}=2000 \mathrm{g}+\mathrm{f}_{\mathrm{r}}=2000+4000\)

\(\mathrm{T}=24000 \mathrm{N}\)

10.Power \(P=T V\)

\(\Rightarrow 60 \times 746=(24000) \mathrm{V}\)

\(\mathrm{V}=\frac{60 \times 746}{24000}=1.865 \approx 1.9 \mathrm{m} / \mathrm{s}\)

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