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Current Question (ID: 10138)

Question:
$\text{The radius of gyration of a uniform solid sphere about a tangent is:}$
Options:
  • 1. $R\sqrt{\frac{2}{3}}$
  • 2. $R\sqrt{\frac{2}{5}}$
  • 3. $R\sqrt{\frac{5}{3}}$
  • 4. $R\sqrt{\frac{7}{5}}$
Solution:
$\text{Hint: Recall the Formula for the radius of gyration.}$ $\text{Step 1: Draw the figure.}$ $\text{The figure shows a solid sphere with radius } R \text{ and a tangent line at distance } R \text{ from the center, with } I_C \text{ representing the moment of inertia about the center.}$ $\text{Step 2: Moment of inertia of a solid sphere about an axis tangent to the solid sphere in terms of the radius of gyration is given as,}$ $I = \frac{2}{5}MR^2 + MR^2 = Mk^2$ $\Rightarrow k = R\sqrt{\frac{7}{5}}$

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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
}