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Current Question (ID: 10257)

Question:
$\text{Three-point masses each of mass } m, \text{ are placed at the vertices of an equilateral triangle of side } a. \text{ The moment of inertia of the system through a mass } m \text{ at } O \text{ and lying in the plane of } COD \text{ and perpendicular to } OA \text{ is:}$
Options:
  • 1. $2ma^2$
  • 2. $\frac{2}{3}ma^2$
  • 3. $\frac{5}{4}ma^2$
  • 4. $\frac{7}{4}ma^2$
Solution:
$\text{M.O.I. of the system about axis COD} = I_O + I_B + I_C$ $= m(0)^2 + m(\frac{a}{2})^2 + ma^2$ $= \frac{ma^2}{4} + ma^2 \text{ (as point mass m at O lies on COD, B lies at distance a/2)}$ $= \frac{5}{4}ma^2 \text{ (and point mass m at C lies at a distance a from the axis COD)}$

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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
}