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Current Question (ID: 10231)

Question:

$\text{The coordinates of the centre of mass of a uniform plate of shape as shown in the figure are:}$

Options:

1. $\frac{L}{2}, \frac{L}{2}$
2. $\frac{5L}{12}, \frac{5L}{12}$
3. $\frac{5L}{3}, \frac{2L}{3}$
4. $\frac{3L}{4}, \frac{L}{2}$

Solution:

\text{Hint: Break the whole system into two parts.} \text{Step 1: Calculate areas of the two parts} \text{Area of small part: } A_1 = \frac{L}{2} \times \frac{L}{2} = \frac{L^2}{4} \text{Area of large part: } A_2 = L \times \frac{L}{2} = \frac{L^2}{2} \text{Since } A_2 = 2A_1, \text{ we have } m_2 = 2m_1 = 2m \text{The centers of mass of the two parts are:} \text{Small part (upper right): } M \cdot \left(\frac{L}{4}, \frac{3L}{4}\right) \text{Large part (lower): } 2M \cdot \left(\frac{L}{2}, \frac{L}{4}\right) \text{Step 2: Calculate center of mass coordinates} x_{cm} = \frac{M \cdot \frac{L}{4} + 2M \cdot \frac{L}{2}}{M + 2M} = \frac{5L}{12} y_{cm} = \frac{M \cdot \frac{3L}{4} + 2M \cdot \frac{L}{4}}{M + 2M} = \frac{5L}{12} \text{Step 3: Final answer} \text{Hence, coordinates of the center of mass = } \left(\frac{5L}{12}, \frac{5L}{12}\right)

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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
}