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Current Question (ID: 19267)

Question:

$\text{A lift of mass } 500 \text{ kg starts moving downwards with an initial speed of } 2 \text{ m/s and accelerates at a rate of } 2 \text{ m/s}^2. \text{ The kinetic energy of the lift when it has moved } 6 \text{ m down, is:}$ $1. \ 3 \text{ J}$ $2. \ 3 \text{ kJ}$ $3. \ 7 \text{ J}$ $4. \ 7 \text{ kJ}$

Options:

1. $3 \text{ J}$
2. $3 \text{ kJ}$
3. $7 \text{ J}$
4. $7 \text{ kJ}$

Solution:

$\text{Hint: } v^2 - u^2 = 2as$ $\text{Step: Find the final kinetic energy of the lift.}$ $\text{Given: } u = 2 \text{ m/s}, a = 2 \text{ m/s}^2, s = 6 \text{ m}$ $\text{By using the third equation of motion we get:}$ $v^2 - u^2 = 2as$ $\text{By multiplying } \frac{m}{2} \text{ both sides we get:}$ $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = 2as \times \frac{m}{2}$ $\Rightarrow \frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mas$ $\Rightarrow \frac{1}{2} \times 500 \times 2^2 + 500 \times 2 \times 6$ $\Rightarrow \frac{1}{2}mv^2 = 1000 + 6000 = 7000 \text{ J} = 7 \text{ kJ}$ $\text{Therefore, the final kinetic energy of the lift is } 7 \text{ kJ.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}