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Current Question (ID: 19087)

Question:

$\text{A particle moves at a constant speed along the circumference of a circle with a radius } R, \text{ subject to a central fictitious force } F \text{ that is inversely proportional to } R^3. \text{ Its time period of revolution will be given by:}$

Options:

1. $T \propto R^2$
2. $T \propto R^{\frac{3}{2}}$
3. $T \propto R^{\frac{5}{2}}$
4. $T \propto R^{\frac{4}{3}}$

Solution:

$\text{Hint: } F = m\omega^2 R$ $\text{Step: Find the time period of the revolution.}$ $\text{Given: The central fictitious force } F \text{ that is inversely proportional to } R^3 \text{ i.e.,}$ $F \propto \frac{1}{R^3} \Rightarrow F = \frac{K}{R^3}$ $\Rightarrow \frac{K}{R^3} = m\omega^2 R \Rightarrow \omega^2 = \frac{K}{mR^4}$ $\Rightarrow \left(\frac{2\pi}{T}\right)^2 = \frac{K}{mR^4} \quad \left[ \omega = \frac{2\pi}{T} \right]$ $\Rightarrow T^2 \propto R^4$ $\Rightarrow T \propto R^2$ $\text{Therefore, the time period of the revolution is proportional to } R^2. \text{ Hence, option (1) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}