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Current Question (ID: 10282)

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:

$\text{1. Relative position vector } r:$ $r = (\text{point of force application}) - (\text{pivot point})$ $r = (2\hat{i} - 3\hat{j} + 6\hat{k}) - (2\hat{i} + 5\hat{j} + 3\hat{k})$ $r = (0\hat{i} - 8\hat{j} + 3\hat{k}) \text{ m}$ $\text{2. Force vector } F:$ $F = (2\hat{i} + 3\hat{j} + 4\hat{k}) \text{ N}$ $\text{3. Cross product } \tau = r \times F:$ $\tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -8 & 3 \\ 2 & 3 & 4 \end{vmatrix}$ $\tau = \hat{i}((-8)(4) - (3)(3)) - \hat{j}((0)(4) - (3)(2)) + \hat{k}((0)(3) - (-8)(2))$ $\tau = \hat{i}(-32 - 9) - \hat{j}(0 - 6) + \hat{k}(0 + 16)$ $\tau = -41\hat{i} + 6\hat{j} + 16\hat{k} \text{ N-m}$ $\text{The final answer is } (-41\hat{i} + 6\hat{j} + 16\hat{k}) \text{ N-m}$

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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
}