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Current Question (ID: 16452)

Question:
$\text{A charge } Q \text{ is enclosed by a Gaussian spherical surface of radius } R. \text{ If the radius is doubled, then the outward electric flux will:}$
Options:
  • 1. $\text{be reduced to half}$
  • 2. $\text{remain the same}$
  • 3. $\text{be doubled}$
  • 4. $\text{increase four times}$
Solution:
$\text{Hint: } \phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$ $\text{Step: Find the outward electric flux if the radius is doubled.}$ $\text{Given,}$ $\text{The radius of the Gaussian surface } = R$ $\text{The charge enclosed inside the Gaussian surface } = Q$ $\text{We know,}$ $\phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$ $\text{Here, total flux doesn’t depend on the size of the Gaussian surface.}$ $\text{Thus, we can say the electric flux depends only on the net enclosed charge by surface.}$ $\text{So, outward electric flux remains the same.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}