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Current Question (ID: 8808)

Question:

$\text{The displacement } x \text{ of a particle moving in one dimension under the action of a constant force is related to time } t \text{ by the equation } t = \sqrt{x} + 3, \text{ where } x \text{ is in meters and } t \text{ is in seconds. What is the displacement of the particle from } t = 0 \text{ s to } t = 6 \text{ s?}$

Options:

1. $0$
2. $12 \text{ m}$
3. $6 \text{ m}$
4. $18 \text{ m}$

Solution:

$\text{Hint: } \Delta x = x_f - x_i$ $\text{Step: Find the displacement of the particle.}$ $\text{The time taken by the particle is given as;}$ $t = \sqrt{x} + 3$ $\Rightarrow x = (t - 3)^2$ $\text{At } t = 0 \text{ s, } x_1 = (0 - 3)^2 = 9 \text{ m}$ $\text{At } t = 6 \text{ s, } x_2 = (6 - 3)^2 = 9 \text{ m}$ $\text{The displacement of the particle is given by;}$ $x_2 - x_1 = 9 - 9 = 0$ $\text{Hence, option (1) is the correct answer.}$

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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
}