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Current Question (ID: 16523)

Question:
$\text{The variation of potential with distance } x \text{ from a fixed point is shown in the figure. The electric field at } x = 13 \text{ m is:}$
Options:
  • 1. $7.5 \text{ V/m}$
  • 2. $-7.5 \text{ V/m}$
  • 3. $5 \text{ V/m}$
  • 4. $-5 \text{ V/m}$
Solution:
$\text{Hint: } E = -\frac{dV}{dx}$ $\text{Step: Find the electric field at } x = 13 \text{ m}$ $\text{As we can see the slope of the graph is constant, so the slope of the graph at point } A \text{ is given by:}$ $E = -\frac{dV}{dx}$ $E = \left[ \frac{30 - 10}{16 - 12} \right] = 5 \text{ V/m}$ $\text{Hence, option (3) is the correct answer.}$

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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
}