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Current Question (ID: 19335)

Question:

$\text{Three particles of masses } 100 \text{ g, } 150 \text{ g, and } 200 \text{ g respectively are placed at the vertices of an equilateral triangle of a side } 0.5 \text{ m long.}$ $\text{What is the position of the centre of mass of three particles?}$

Options:

1. $\left( \frac{5}{18}, \frac{1}{3\sqrt{3}} \right)$
2. $\left( \frac{1}{4}, 0 \right)$
3. $\left( 0, \frac{1}{4} \right)$
4. $\left( \frac{1}{3\sqrt{3}}, \frac{5}{18} \right)$

Solution:

$\text{Hint: Position of centre of mass, } x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} \text{ and } y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}$ $\text{Step 1: Find the } x\text{-coordinate of the centre of mass.}$ $X = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$ $= \frac{100 (0) + 150 (0.5) + 200 (0.25)}{(100 + 150 + 200)} \text{ g-m}$ $= \frac{75 + 50}{450} \text{ m} \Rightarrow \frac{125}{450} \text{ m} = \frac{5}{18} \text{ m}$ $\text{Step 2: Find the } y\text{-coordinate of the centre of mass.}$ $y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}$ $Y = \frac{[100 (0) + 150 (0) + 200 (0.25 \sqrt{3})]}{450} \text{ g-m}$ $= \frac{50 \sqrt{3}}{450} \text{ m} \Rightarrow \frac{\sqrt{3}}{9} \text{ m} = \frac{1}{3\sqrt{3}} \text{ m}$ $\text{The centre of mass } C \text{ is shown in the figure. Note that it is not the geometric centre of the triangle OAB.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}