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Current Question (ID: 20261)

Question:
$\text{Find the equivalent resistance between points } A \text{ and } B \text{ in the given circuit.}$
Options:
  • 1. $6.4 \, \Omega$
  • 2. $4 \, \Omega$
  • 3. $3.2 \, \Omega$
  • 4. $8 \, \Omega$
Solution:
$\text{Hint: Use symmetry to simplify the circuit.}$ $\text{Step: Find the equivalent resistance across } A \text{ and } B.$ $\text{The equivalent circuit diagram of the resistance is shown in the figure below;}$ $\text{All } 4 \, \Omega \text{ resistance forms the Wheatstone Bridge and the equivalent}$ $\text{resistance of the Wheatstone Bridge configuration is } 4 \, \Omega.$ $\left[ \frac{8 \times 8}{8 + 8} = 4 \, \Omega \right]$ $\text{The equivalent resistance across } A \text{ and } B \text{ is given by;}$ $R_{AB} = \frac{4 \times 16}{16 + 4} = \frac{64}{20} = \frac{16}{5} \, \Omega$ $\text{Hence, option (3) is the correct answer.}$

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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
}