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Current Question (ID: 16628)

Question:
$\text{Variation of current passing through a conductor with the voltage applied across its ends varies is shown in the diagram below. If the resistance } (R) \text{ is determined at points } A, B, C \text{ and } D, \text{ we will find that:}$
Options:
  • 1. $R_C = R_D$
  • 2. $R_B > R_A$
  • 3. $R_C > R_B$
  • 4. $\text{None of these}$
Solution:
$\text{From the curve, it is clear that slopes at points } A, B, C, \text{ and } D \text{ have the following order: } A > B > C > D.$ $\text{Also, the resistance at any point equals the slope of the V-I curve.}$ $\text{So, the order of resistance at the three points will be } R_A > R_B > R_C > R_D.$

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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
}