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Current Question (ID: 19442)

Question:
$\text{A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius } R_e. \text{ By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that is become } \sqrt{\frac{3}{2}} \text{ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is } R. \text{ Value of } R \text{ is ?}$
Options:
  • 1. $2R_e$
  • 2. $2.5R_e$
  • 3. $3R_e$
  • 4. $4R_e$
Solution:
$\text{Hint: Apply the conservation of energy.}$ $V = \sqrt{\frac{3}{2}} V_0$ $V_0 = \sqrt{\frac{GM}{R_e}}$ $-\frac{GMm}{R_e} + \frac{1}{2} mv^2 = -\frac{GMm}{R_{\text{max}}} + \frac{1}{2} mv'^2$ $VR_e = V'R_{\text{max}}$ $Solving \ (i) \ and \ (ii)$ $R_{\text{max}} = 3R_e$

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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
}