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Current Question (ID: 8901)

Question:
$\text{A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:}$
Options:
  • 1. $\text{B}$
  • 2. $\text{C}$
  • 3. $\text{D}$
  • 4. $\text{A}$
Solution:
\text{Hint: } v_x = \frac{dx}{dt} \text{Explanation: The particle has a maximum instantaneous velocity at that point at which its slope is maximum.} \text{Therefore, } v_{\text{max}} = \frac{dx}{dt} = \text{maximum slope.} \text{From the figure, it is obvious that at point C, the slope is maximum, hence at this point velocity is maximum.} \text{Hence, option (2) 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
}