Import Question JSON

Current Question (ID: 9666)

Question:

$\text{An object moves at a constant speed along a circular path in a horizontal XY plane with its centre at the origin. When the object is at } x = -2 \text{ m, its velocity is } -(4 \text{ m/s})\hat{j}\text{. What is the object's acceleration when it is at } y = 2 \text{ m?}$

Options:

1. $-8 \text{ m/s}^2 \hat{j}$
2. $-8 \text{ m/s}^2 \hat{i}$
3. $-4 \text{ m/s}^2 \hat{j}$
4. $-4 \text{ m/s}^2 \hat{i}$

Solution:

\text{For uniform circular motion, the centripetal acceleration is always directed towards the center of the circle.} \text{Magnitude of the acceleration } = \frac{v^2}{r} = \frac{4 \times 4}{2} = 8 \text{ m/s}^2 \text{When the object is at } y = 2 \text{ m, the acceleration is directed towards the center (origin), which is in the negative y-direction.} \vec{a} = -a\hat{r} = -8\hat{j} \text{ m/s}^2

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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
}