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Current Question (ID: 19487)

Question:
$\text{A body of mass } m \text{ is moving in a circular orbit of radius } R \text{ about a planet of mass } M. \text{ At some instant, it splits into two equal masses.}$ $\text{The first mass moves in a circular orbit of radius } \frac{R}{2}, \text{ and the other mass, in a circular orbit of radius } \frac{3R}{2}. \text{ The difference between the final and initial total energies is:}$
Options:
  • 1. $+\frac{Gm}{6R}$
  • 2. $-\frac{GMm}{2R}$
  • 3. $-\frac{GMm}{6R}$
  • 4. $\frac{GMm}{2R}$
Solution:
$\text{Hint: } U = -\frac{GMm}{R}$

Import JSON File

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
}