Import Question JSON

Current Question (ID: 10162)

Question:

$\text{A force } \vec{F} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \text{ N is acting at point } (2\text{ m}, -3\text{ m}, 6\text{ m}). \text{ Find the torque of this force about a point whose position vector is } (2\hat{i} + 5\hat{j} + 3\hat{k}) \text{ m.}$

Options:

1. $\vec{\tau} = (-17\hat{i} + 6\hat{j} + 4\hat{k}) \text{ N-m}$
2. $\vec{\tau} = (-17\hat{i} + 6\hat{j} - 4\hat{k}) \text{ N-m}$
3. $\vec{\tau} = (17\hat{i} - 6\hat{j} + 4\hat{k}) \text{ N-m}$
4. $\vec{\tau} = (-41\hat{i} + 6\hat{j} + 16\hat{k}) \text{ N-m}$

Solution:

\text{The torque } \tau \text{ is calculated as } \tau = \vec{r} \times \vec{F}. \text{First, find } \vec{r} = \vec{r}_{\text{force}} - \vec{r}_{\text{pivot}} = (2\hat{i} - 3\hat{j} + 6\hat{k}) - (2\hat{i} + 5\hat{j} + 3\hat{k}) = -8\hat{j} + 3\hat{k} \text{ m} \text{Then, calculate the cross product:} \tau = (-8\hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) \tau = (-8)(2)(\hat{j} \times \hat{i}) + (-8)(4)(\hat{j} \times \hat{k}) + (3)(2)(\hat{k} \times \hat{i}) + (3)(3)(\hat{k} \times \hat{j}) \tau = -16(-\hat{k}) - 32(\hat{i}) + 6(\hat{j}) + 9(-\hat{i}) \tau = 16\hat{k} - 32\hat{i} + 6\hat{j} - 9\hat{i} \tau = (-32 - 9)\hat{i} + 6\hat{j} + 16\hat{k} \tau = -41\hat{i} + 6\hat{j} + 16\hat{k} \text{ N·m}

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
}