Import Question JSON

Current Question (ID: 20148)

Question:
$\text{Pick the correct graph between potential } V \text{ at distance } r \text{ from centre}$ $\text{for the uniformly charged spherical shell of radius } R.$
Options:
  • 1. $\text{Graph 1}$
  • 2. $\text{Graph 2}$
  • 3. $\text{Graph 3}$
  • 4. $\text{Graph 4}$
Solution:
$\text{Hint: } V = \frac{KQ}{R}$ $\text{Step: Find the variation of potential as a function of } r \text{ for a}$ $\text{spherical shell.}$ $\text{For a uniformly charged spherical shell of radius } R, \text{ the potential } V$ $\text{at a distance } r \text{ from the centre behaves differently depending on}$ $\text{whether } r \text{ is inside or outside the shell.}$ $\text{The potential is given as } V = - \int E \cdot dl$ $\text{The potential at the centre of the spherical shell of the radius } R \text{ is}$ $\text{constant is given by;}$ $V = \frac{KQ}{R} \text{ for } r \leq R$ $\text{The potential at outside the surface of the spherical shell varies}$ $V \propto \frac{1}{r}.$ $V = \frac{KQ}{r} \text{ for } r > R$ $\text{Therefore, the graph should show a flat line inside the shell and a}$ $\text{hyperbolic decay outside the shell as shown in the figure below;}$ $\text{Hence, option (4) is the correct answer.}$

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
}