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Question:
$\text{Two blocks of masses } 10 \text{ kg and } 30 \text{ kg are placed on the same straight line with coordinates } (0, 0) \text{ cm and } (x, 0) \text{ cm respectively.}$ $\text{The block of } 10 \text{ kg is moved on the same line through a distance of } 6 \text{ cm towards the other block.}$ $\text{The distance through which the block of } 30 \text{ kg must be moved to keep the position of centre of mass of the system unchanged is:}$
Options:
  • 1. $4 \text{ cm towards the } 10 \text{ kg block}$
  • 2. $2 \text{ cm away from the } 10 \text{ kg block}$
  • 3. $2 \text{ cm towards the } 10 \text{ kg block}$
  • 4. $4 \text{ cm away from the } 10 \text{ kg block}$
Solution:
$\text{Hint: } x_{\text{COM}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ $\text{Step 1: Find the initial position of the centre of mass.}$ $x_{\text{COM}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{(10 \times 0) + (30 \times x)}{10 + 30} = \frac{30x}{40} = \frac{3x}{4}$ $\text{Step 2: Find the distance through which the block of } 30 \text{ kg must be moved.}$ $\text{Now, the } 10 \text{ kg block is moved } 6 \text{ cm towards the } 30 \text{ kg block, so its new position becomes } x_1' = 6 \text{ cm.}$ $\text{Let's assume the } 30 \text{ kg block moves by a distance } d. \text{ Its new position will be } x_2' = x + d.$ $\text{The center of mass should remain unchanged:}$ $\frac{10 \times 6 + 30 \times (x + d)}{10 + 30} = \frac{3x}{4}$ $\frac{60 + 30(x + d)}{40} = \frac{3x}{4}$ $60 + 30(x + d) = 30x$ $\Rightarrow d = -2 \text{ cm}$ $\text{The } 30 \text{ kg block must be moved } 2 \text{ cm towards the } 10 \text{ kg block to keep the centre of mass unchanged.}$ $\text{Hence, option } (3) \text{ is the correct answer.}$

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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
}