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Current Question (ID: 9552)

Question:
$\text{A body starts moving from rest on a horizontal ground such that the position vector of the body with respect to its starting point is given by } \vec{r} = 2t\hat{i} + 3t^2\hat{j}. \text{ The equation of the trajectory of the body is:}$
Options:
  • 1. $y = 1.5x$
  • 2. $y = 0.75x^2$
  • 3. $y = 1.5x^2$
  • 4. $y = 0.45x^2$
Solution:
$\text{Hint: Correlate the } x \text{ and } y \text{ coordinate of the body.}$ $\text{Step: Find the equation of the trajectory of the body.}$ $\text{Given, } \vec{r} = 2t\hat{i} + 3t^2\hat{j}$ $\Rightarrow x = 2t \text{ & } y = 3t^2$ $\Rightarrow t = \frac{x}{2}$ $\text{Putting it into } y = 3t^2$ $\Rightarrow y = 3 \times \frac{x^2}{4}$ $\Rightarrow y = 0.75x^2$ $\text{Hence, option (2) is the correct answer.}$

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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
}