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Current Question (ID: 19338)

Question:

$\text{A square shaped hole of side } l = \frac{a}{2} \text{ is carved out at a distance } d = \frac{a}{2} \text{ from the centre } 'O' \text{ of a uniform circular disk of radius } a. \text{ If the distance of the centre of mass of the remaining portion from } O \text{ is } -\frac{a}{x} \text{ value of } x \text{ (to the nearest integer) is:}$

Options:

1. $12$
2. $23$
3. $45$
4. $76$

Solution:

$\text{Hint: } X_{\text{COM}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ $X_{\text{com}} = \frac{m_1x_1 - m_2x_2}{m_1 - m_2}$ $\text{where:}$ $m_1 = \text{mass of complete disc}$ $m_2 = \text{removed mass}$ $\text{Let } \sigma = \text{surface mass density of disc material}$ $w.r.t \ 'O': \ X_{\text{com}} = \frac{\sigma \pi a^2 (O) - \sigma \cdot \frac{a^2}{4} \cdot d}{\sigma \pi a^2 - \sigma \frac{a^2}{4}} = \frac{-\frac{a^2}{4} \cdot d}{\pi a^2 - \frac{a^2}{4}} = \frac{-d}{4\pi - 1}$ $\text{So, } X = 2(4\pi - 1) = (8\pi - 2) = 23.12$ $\text{So, nearest integer value of } X = 23$

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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
}