Import Question JSON

Current Question (ID: 19325)

Question:

$\text{A uniform rectangular thin sheet } ABCD \text{ of mass } M \text{ has length } a \text{ and breadth } b. \text{ If the shaded portion } HBGO \text{ is removed, the coordinates of the centre-of-mass of the remaining portion will be:}$

Options:

1. $\left( \frac{5a}{3}, \frac{5b}{3} \right)$
2. $\left( \frac{2a}{3}, \frac{2b}{3} \right)$
3. $\left( \frac{5a}{12}, \frac{5b}{12} \right)$
4. $\left( \frac{3a}{4}, \frac{3b}{4} \right)$

Solution:

$\text{Hint: } x_{\text{com}} = \frac{M_1x_1 - M_2x_2}{M_1 - M_2}$ $\text{Step: Find the coordinates of the centre of mass of the remaining portion.}$ $\text{The } x\text{-coordinate of centre of mass:}$ $x_{\text{com}} = \frac{M_1x_1 - M_2x_2}{M_1 - M_2}$ $x_{\text{com}} = \frac{M \times \frac{a}{2} - \frac{M}{4} \times \frac{3a}{4}}{M - \frac{M}{4}}$ $\Rightarrow x_{\text{com}} = \frac{5a}{12}$ $\text{The } y\text{-coordinate of centre of mass:}$ $y_{\text{com}} = \frac{M_1y_1 - M_2y_2}{M_1 - M_2}$ $y_{\text{com}} = \frac{M \times \frac{b}{2} - \frac{M}{4} \times \frac{3b}{4}}{M - \frac{M}{4}}$ $\Rightarrow y_{\text{com}} = \frac{5b}{12}$ $\text{Therefore, the coordinates of the centre of mass of the remaining portion is } \left( \frac{5a}{12}, \frac{5b}{12} \right).$ $\text{Hence, option (3) is the correct answer.}$

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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
}