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Current Question (ID: 9725)

Question:
$\text{A 0.5 kg body experiences a force } F = (2 + 3x^2) \text{ N, where } x \text{ in metres is the displacement from the origin. If it is released to move along the } X\text{-axis from the origin, then its initial acceleration is:}$
Options:
  • 1. $2 \text{ m/s}^2$
  • 2. $10 \text{ m/s}^2$
  • 3. $4 \text{ m/s}^2$
  • 4. $\text{zero}$
Solution:
$\text{Hint: } F = ma$ $\text{Step 1: Identify initial position } x = 0.$ $\text{Step 2: Use 2}^{\text{nd}} \text{ law of motion.}$ $F = ma$ $\text{At } x = 0: F = 2 + 3(0)^2 = 2 \text{ N}$ $\text{Using } F = ma: 2 = 0.5 \times a$ $\therefore a = \frac{2}{0.5} = 4 \text{ m/s}^2$

Import JSON File

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
}