Import Question JSON

Current Question (ID: 20061)

Question:

$\text{Given below are two statements:}$ $\text{Assertion (A):}$ \text{When an electric dipole is completely enclosed by a closed Gaussian surface, the total electric flux through the surface is zero.}$ $\text{Reason (R):}$ \text{The net charge enclosed within the surface is zero.}$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True but (R) is False.}$
4. $\text{Both (A) and (R) are False.}$

Solution:

$\text{Hint: } \phi = \frac{q_{\text{in}}}{\varepsilon_0} \text{ and } q_{\text{in}} = 0 \text{ (inside surface)}$ $\text{Explanation: As Gauss's theorem states, the total electric flux through a closed surface is proportional to the net charge enclosed divided by the permittivity of free space. When an electric dipole is enclosed in a closed Gaussian surface, the total electric flux through the surface is zero because the net charge inside the enclosed surface, which is the sum of the positive and negative charges of the dipole, is zero. This satisfies the necessary condition for zero net flux according to Gauss's Law. Therefore, both (A) and (R) are True and (R) is the correct explanation of (A). Hence, option (1) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}