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Current Question (ID: 10101)

Question:

$\text{Three masses are placed on the x-axis: 300 g at the origin, 500 g at } x = 40 \text{ cm, and 400 g at } x = 70 \text{ cm. The distance of the center of mass from the origin is:}$

Options:

1. $40 \text{ cm}$
2. $45 \text{ cm}$
3. $50 \text{ cm}$
4. $30 \text{ cm}$

Solution:

$\text{NCERT Reference: Class XI, part 1, pg 145}$ $\text{Hint: } x_{CM} = \frac{\sum m_i x_i}{\sum m_i}$ $\text{Using the center of mass formula:}$ $x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$ $\text{Substituting the given values:}$ $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400}$ $= \frac{0 + 20000 + 28000}{1200}$ $= \frac{48000}{1200}$ $= 40 \text{ cm}$ $\text{Therefore, the distance of the center of mass from the origin is 40 cm.}$

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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
}