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Current Question (ID: 19480)

Question:

$\text{Two identical particles each of mass } m, \text{ move in circular path due to their own mutual gravitational force.}$ $\text{Find the velocity of the particle if the radius of circular path is } a$

Options:

1. $\sqrt{\frac{4Gm}{a}}$
2. $\sqrt{\frac{Gm}{2a}}$
3. $\sqrt{\frac{2Gm}{a}}$
4. $\sqrt{\frac{Gm}{4a}}$

Solution:

$\text{Hint: } F = G \frac{m_1 m_2}{r^2}$ $\frac{mv^2}{a} = \frac{Gm^2}{4a^2} \Rightarrow v = \sqrt{\frac{Gm}{4a}}$

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
}