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Current Question (ID: 10103)

Question:

$\text{The coordinates of the position of masses } m_1 = 7 \text{ gm}, m_2 = 4 \text{ gm}, m_3 = 10 \text{ gm are } \vec{r}_1 = (\hat{i} + 5\hat{j} - 3\hat{k}), \vec{r}_2 = (2\hat{i} + 5\hat{j} + 7\hat{k}), \vec{r}_3 = (3\hat{i} + 3\hat{j} - \hat{k}) \text{ respectively in cm. The position of the centre of mass of the system would be:}$

Options:

1. $\left(\frac{15}{7}, \frac{85}{17}, \frac{1}{7}\right) \text{ cm}$
2. $\left(\frac{15}{7}, -\frac{85}{17}, \frac{1}{7}\right) \text{ cm}$
3. $\left(\frac{15}{7}, \frac{85}{21}, -\frac{1}{7}\right) \text{ cm}$
4. $\left(\frac{15}{7}, \frac{85}{21}, \frac{7}{3}\right) \text{ cm}$

Solution:

$\text{Hint: Recall the formula for center of mass in vector form}$ $\text{Step 1: Write general form for center of mass}$ $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$ $\text{Step 2: Calculate the } \vec{r}_{cm}\text{:}$ $\vec{r}_{cm} = \frac{7(\hat{i} + 5\hat{j} - 3\hat{k}) + 4(2\hat{i} + 5\hat{j} + 7\hat{k}) + 10(3\hat{i} + 3\hat{j} - \hat{k})}{7 + 4 + 10}$ $\text{Expanding the numerator:}$ $= \frac{7\hat{i} + 35\hat{j} - 21\hat{k} + 8\hat{i} + 20\hat{j} + 28\hat{k} + 30\hat{i} + 30\hat{j} - 10\hat{k}}{21}$ $= \frac{45\hat{i} + 85\hat{j} - 3\hat{k}}{21}$ $= \frac{15}{7}\hat{i} + \frac{85}{21}\hat{j} - \frac{1}{7}\hat{k}$ $\text{Hence, option (3) is the correct answer.}$

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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
}