Import Question JSON

Current Question (ID: 20018)

Question:

$\text{A long cylindrical shell is uniformly charged such that the upper half of its curved surface carries a positive surface charge density } \sigma, \text{ while the lower half carries an equal negative surface charge density } -\sigma. \text{ Which of the following schematic diagrams (not drawn to scale) most accurately represents the resulting pattern of electric field lines around the cylinder?}$

Options:

1. $\text{Diagram 1}$
2. $\text{Diagram 2}$
3. $\text{Diagram 3}$
4. $\text{Diagram 4}$

Solution:

$\text{Hint: Electric field lines start from positive charges and terminate at negative charges.}$ $\text{Step: Analyse each pattern of electric field lines around the cylinder and find the correct answer.}$ $(1) \text{ Outside the cylinder the field lines leave the positively charged upper semicircle and terminate on the negatively charged lower semicircle.}$ $\text{Hence, pattern (1) is correct.}$ $(2) \text{ The electric field lines cannot be straight line in the case of cylindrical shell.}$ $\text{Hence, pattern (2) is incorrect.}$ $(3) \text{ Electric field is absent in the left and right regions of cylindrical shell.}$ $\text{Hence, pattern (3) is incorrect.}$ $(4) \text{ Electric field is originating from negative side of cylindrical shell.}$ $\text{Hence, pattern (4) is incorrect.}$ $\text{Hence, option (1) is the correct answer.}$

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
}