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Current Question (ID: 20044)

Question:
$\text{A long cylindrical volume contains a uniformly distributed charge of density } \rho \text{ cm}^{-3}. \text{ The electric field inside the cylindrical volume at a distance } x = \frac{2\varepsilon_0}{\rho} \text{ m from its axis is:}$
Options:
  • 1. $1 \text{ Vm}^{-1}$
  • 2. $2 \text{ Vm}^{-1}$
  • 3. $3 \text{ Vm}^{-1}$
  • 4. $4 \text{ Vm}^{-1}$
Solution:
$\text{Hint: } \oint \vec{E} \cdot d\vec{S} = \frac{Q}{\varepsilon_0}$ $\text{Step: Find the electric field at the given distance. Apply the Gauss law.}$ $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{in}}}{\varepsilon_0}$ $E(2\pi x h) = \rho \times \frac{\pi x^2 h}{\varepsilon_0}$ $E = \frac{\rho x}{2\varepsilon_0} = \frac{\rho}{2\varepsilon_0} \times \frac{2\varepsilon_0}{\rho} = 1$

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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
}