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Current Question (ID: 19375)

Question:
$\text{The radius of gyration for the uniform solid sphere of radius } 5 \text{ cm}$ $\text{about the axis } PQ, \text{ as shown in the figure is:}$
Options:
  • 1. $5 \text{ cm}$
  • 2. $10 \text{ cm}$
  • 3. $\sqrt{110} \text{ cm}$
  • 4. $\sqrt{90} \text{ cm}$
Solution:
$\text{Hint: Use the parallel axis theorem.}$ $\text{Step: Find the radius of gyration for the uniform solid sphere.}$ $\text{By using the parallel axis theorem, the moment of inertia of the}$ $\text{uniform solid sphere is given by;}$ $I = I_{\text{cm}} + Md^2 \quad \cdots (1)$ $I = MK^2 \quad \cdots (2)$ $\text{From equation (1) and (2) we get;}$ $MK^2 = M \left( \frac{2}{5} R^2 + d^2 \right) \quad \left[ I_{\text{cm}} = \frac{2}{5} MR^2 \right]$ $\Rightarrow K^2 = \frac{2}{5} \times 25 + 100 \quad \left[ d = 10 \text{ cm} \right]$ $\Rightarrow K = \sqrt{110} \text{ cm}$ $\text{Therefore, the radius of gyration for the uniform solid sphere is}$ $\sqrt{110} \text{ cm.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}