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Current Question (ID: 19272)

Question:

$\text{A lift with a mass of } 1400 \text{ kg moves upwards at a constant velocity of } 3 \text{ m/s, and experiences a frictional force of } 2000 \text{ N.}$ $\text{What is the power of the motor driving the lift? (take } g = 10 \text{ m/s}^2)$

Options:

1. $48 \text{ kW}$
2. $24 \text{ kW}$
3. $32 \text{ kW}$
4. $42 \text{ kW}$

Solution:

$\text{Hint: } P = \vec{F} \cdot \vec{v}$ $\text{Step: Find the power of the motor driving the lift.}$ $\text{The power of the motor is given by:}$ $P = \vec{F} \cdot \vec{v}$ $\Rightarrow P = (mg + f_r) \cdot v$ $\Rightarrow P = (1400 \times 10 + 2000) \cdot 3$ $\Rightarrow P = 16000 \times 3 = 48000 \text{ watt} = 48 \text{ kW}$ $\text{Therefore, the power of the motor driving the lift is } 48 \text{ kW.}$ $\text{Hence, option (1) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}