Import Question JSON

Current Question (ID: 20104)

Question:

$\text{The figure shows two point charges } +Q \text{ and } -Q \text{ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If } \sigma_1 \text{ is the surface charge on the inner surface and } Q_1 \text{ net charge on it and } \sigma_2 \text{ the surface charge on the outer surface and } Q_2 \text{ net charge on it then:}$

Options:

1. $\sigma_1 \neq 0, Q_1 = 0, \text{ and } \sigma_2 \neq 0, Q_2 = 0$
2. $\sigma_1 = 0, Q_1 = 0, \text{ and } \sigma_2 = 0, Q_2 = 0$
3. $\sigma_1 \neq 0, Q_1 \neq 0, \text{ and } \sigma_2 \neq 0, Q_2 \neq 0$
4. $\sigma_1 \neq 0, Q_1 = 0, \text{ and } \sigma_2 = 0, Q_2 = 0$

Solution:

$\text{Hint: Inside the cavity, the net charge is zero.}$ $\text{Explanation: When two opposite charges } +Q \text{ and } -Q \text{ are placed inside the cavity of a conducting spherical shell, they induce charges on the inner surface of the cavity to ensure the electric field inside the conductor remains zero. Since the total charge inside the cavity is zero, the net induced charge on the inner cavity surface is also zero. As a result, no charge is needed on the outer surface to balance anything, so it also carries zero net charge. Therefore, both the inner surface of the cavity and the outer surface of the shell have zero net charge.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}