Import Question JSON

Current Question (ID: 10326)

Question:
$\text{The moment of inertia of a horizontal ring about its vertical axis through the centre is } mR^2. \text{ The moment of inertia about its tangent parallel to the plane is:}$
Options:
  • 1. $\frac{3mR^2}{2}$
  • 2. $\frac{mR^2}{4}$
  • 3. $\frac{mR^2}{2}$
  • 4. $\frac{3mR^2}{4}$
Solution:
\text{1. From perpendicular axis theorem:} \text{The moment of inertia of a ring about its diameter } I_d \text{ is half of its moment} \text{of inertia about the axis perpendicular to its plane passing through the center.} I_c = mR^2 \text{So, } I_d = \frac{mR^2}{2} \text{2. From parallel axis theorem:} \text{The moment of inertia about a tangent parallel to the plane is found by} \text{adding the moment of inertia about the parallel diameter } I_d \text{ to } mR^2 \text{where R is the distance from the diameter to the tangent.} \text{So, } I_{\text{tangent}} = I_d + mR^2 = \frac{mR^2}{2} + mR^2 = \frac{3mR^2}{2}

Import JSON File

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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
}