Import Question JSON

Current Question (ID: 16713)

Question:
$\text{The potential difference } V_A - V_B \text{ between the points } A \text{ and } B \text{ in the given figure is:}$ $\begin{array}{c} V_A \xrightarrow{I=2\,\text{A}} 2\,\Omega \quad 3\,\text{V} \quad 1\,\Omega \xrightarrow{} V_B \end{array}$
Options:
  • 1. $-3\,\text{V}$
  • 2. $+3\,\text{V}$
  • 3. $+6\,\text{V}$
  • 4. $+9\,\text{V}$
Solution:
$\text{Hint: Use Kirchoff's voltage law in the given circuit.}$ $\text{Step: Find the potential difference } V_A - V_B \text{ between the points } A \text{ and } B$ $\text{Use Kirchoff's voltage law in the given circuit we get:}$ $V_B = V_A - (2 \times 2) - 3 - (2 \times 1)$ $\Rightarrow V_A - V_B = 9\,\text{V}$ $\text{Hence, option (4) is the correct answer.}$

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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
}