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Current Question (ID: 20280)

Question:
$\text{Space between two concentric conducting spheres of radii } a \text{ and } b \ (b > a) \text{ is filled with a medium of resistivity } \rho. \text{ The resistance between the two spheres will be:}$
Options:
  • 1. $\frac{\rho}{2\pi} \left( \frac{1}{a} + \frac{1}{b} \right)$
  • 2. $\frac{\rho}{4\pi} \left( \frac{1}{a} + \frac{1}{b} \right)$
  • 3. $\frac{\rho}{2\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$
  • 4. $\frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$
Solution:
$\text{Hint: Resistance of spherical shell is } R = \frac{\rho}{4\pi} \left[ \frac{b-a}{ab} \right]$

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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
}