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Current Question (ID: 19463)

Question:

$\text{Two planets } A \text{ and } B \text{ of equal masses are having their periods of revolution } T_A \text{ and } T_B \text{ such that } T_A = 2T_B. \text{ These planets are revolving in the circular orbits of radii } r_A \text{ and } r_B \text{ respectively. Which of the following would be the correct relationship of their orbits?}$

Options:

1. $2r_A^2 = r_B^2$
2. $r_A^3 = 2r_B^3$
3. $r_A^3 = 4r_B^3$
4. $T_A^2 - T_B^2 = \frac{\pi^2}{GM} \left( r_B^3 - 4r_A^3 \right)$

Solution:

$\text{Hint: } T^2 \propto r^3$ $\text{Step: Find the relation between the radii of the planet's orbit.}$ $\text{The formula for a planet's time period is given by:}$ $T^2 = \frac{4\pi^2 r^3}{GM}$ $\Rightarrow T^2 \propto r^3$ $\Rightarrow \left( \frac{T_A}{T_B} \right)^2 = \left( \frac{r_A}{r_B} \right)^3$ $\Rightarrow \left( \frac{2}{1} \right)^2 = \left( \frac{r_A}{r_B} \right)^3$ $\Rightarrow r_A^3 = 4r_B^3$ $\text{Hence option (3) is the correct answer.}$

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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
}