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Current Question (ID: 9836)

Question:

$\text{Forces acting on a particle have magnitudes of 14, 7, and 7 N and act in the direction of vectors } 6\hat{i} + 2\hat{j} + 3\hat{k}, 3\hat{i} - 2\hat{j} + 6\hat{k}, 2\hat{i} - 3\hat{j} - 6\hat{k} \text{ respectively. The forces remain constant while the particle is displaced from point } A: (2, 1, -3) \text{ to } B: (5, 1, 1). \text{ The coordinates are specified in meters. The work done equal to:}$

Options:

1. $75 \text{ J}$
2. $55 \text{ J}$
3. $85 \text{ J}$
4. $65 \text{ J}$

Solution:

$\text{Work done is calculated using } W = \vec{F} \cdot \vec{d}$ $\text{First, find the displacement vector } \vec{d}:$ $\vec{d} = (5-2)\hat{i} + (1-1)\hat{j} + (1-(-3))\hat{k} = 3\hat{i} + 4\hat{k}$ $\text{Next, find the resultant force } \vec{F}:$ $\text{The three forces are:}$ $\vec{F_1} = 14 \cdot \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{6^2 + 2^2 + 3^2}} = 14 \cdot \frac{6\hat{i} + 2\hat{j} + 3\hat{k}}{7} = 12\hat{i} + 4\hat{j} + 6\hat{k}$ $\vec{F_2} = 7 \cdot \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{\sqrt{3^2 + (-2)^2 + 6^2}} = 7 \cdot \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{7} = 3\hat{i} - 2\hat{j} + 6\hat{k}$ $\vec{F_3} = 7 \cdot \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{\sqrt{2^2 + (-3)^2 + (-6)^2}} = 7 \cdot \frac{2\hat{i} - 3\hat{j} - 6\hat{k}}{7} = 2\hat{i} - 3\hat{j} - 6\hat{k}$ $\text{Resultant force:}$ $\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = (12\hat{i} + 4\hat{j} + 6\hat{k}) + (3\hat{i} - 2\hat{j} + 6\hat{k}) + (2\hat{i} - 3\hat{j} - 6\hat{k})$ $\vec{F} = 17\hat{i} + \hat{j} + 6\hat{k}$ $\text{Calculate work done:}$ $W = \vec{F} \cdot \vec{d} = (17\hat{i} + \hat{j} + 6\hat{k}) \cdot (3\hat{i} + 4\hat{k}) = 17 \times 3 + 1 \times 0 + 6 \times 4 = 51 + 24 = 75 \text{ J}$

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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
}