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Current Question (ID: 16729)

Question:
$\text{A capacitor } C \text{ is charged with the help of a resistance } R \text{ as shown in the figure, Variation of } (V_R + V_C) \text{ with time } t \text{ is correctly shown in which of the options?}$ $\text{(} V_R \text{ and } V_C \text{ are instantaneous potential drops across } R \text{ and } C \text{ respectively).}$
Options:
  • 1. $V_0 \text{ decreasing with } t$
  • 2. $V_0 \text{ increasing with } t$
  • 3. $V_0 \text{ constant with } t$
  • 4. $V_0 \text{ zero with } t$
Solution:
$\text{Hint: Apply Kirchhoff's voltage law.}$ $\text{Step: Find the value of } (V_R + V_C).$ $\text{Apply Kirchhoff's voltage law.}$ $V_A + V_0 - V_C - V_R = V_A$ $V_0 = V_C + V_R$ $\therefore (V_R + V_C) \text{ is constant.}$

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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
}