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Current Question (ID: 19522)

Question:
$\text{Correct formula for height of a satellite from earth's surface is -}$
Options:
  • 1. $\left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R$
  • 2. $\left( \frac{T^2 R^2 g}{4 \pi} \right)^{1/2} - R$
  • 3. $\left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{1/3} - R$
  • 4. $\left( \frac{T^2 R^2 g}{4 \pi} \right)^{-1/3} + R$
Solution:
$\text{The correct formula for the height of a satellite from the earth's surface is derived from the orbital mechanics equations.}$ $\text{The formula is:}$ $h = \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R$ $\text{where } T \text{ is the orbital period, } R \text{ is the radius of the earth, and } g \text{ is the acceleration due to gravity.}$

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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
}