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Current Question (ID: 16576)

Question:
$\text{The insulation property of air breaks down at } E = 3 \times 10^6 \text{ V/m.}$ $\text{The maximum charge that can be given to a sphere of diameter } 5 \text{ m is approximately:}$
Options:
  • 1. $2 \times 10^{-5} \text{ C}$
  • 2. $2.2 \times 10^{-4} \text{ C}$
  • 3. $2 \times 10^{-3} \text{ C}$
  • 4. $3 \times 10^{-3} \text{ C}$
Solution:
$\text{For maximum charge,}$ $r = \frac{5}{2} = 2.5 \text{ m}$ $\text{so, } 3 \times 10^6 = 9 \times 10^9 \times \frac{q_{\text{max}}}{(r)^2}$ $\Rightarrow q_{\text{max}} = \frac{3 \times 10^6 \times r^2}{9 \times 10^9} = \frac{r^2}{3 \times 10^3}$ $= \frac{2.5 \times 2.5 \times 10^{-3}}{3}$ $= 2 \times 10^{-3} \text{ C}$

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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
}