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Current Question (ID: 10137)

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:

$\text{M.O.I. of the system about axis COD} = I_O + I_B + I_C$ $= m(0)^2 + m\left(\frac{a}{2}\right)^2 + ma^2$ $\text{From the geometry of the equilateral triangle:}$ $\text{Point mass } m \text{ at } O \text{ lies on COD, so distance = 0}$ $\text{Point mass } m \text{ at } B \text{ lies at distance } \frac{a}{2} \text{ from axis COD}$ $\text{Point mass } m \text{ at } C \text{ lies at distance } a \text{ from axis COD}$ $= \frac{ma^2}{4} + ma^2 \text{ (as point mass } m \text{ at } O \text{ lies on COD, } B \text{ lies at distance } \frac{a}{2})$ $= \frac{5}{4}ma^2 \text{ (and point mass } m \text{ at } C \text{ lies at a distance } a \text{ from the axis COD)}$

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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
}