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Current Question (ID: 16632)

Question:

$\text{The equivalent resistance between } A \text{ and } B \text{ for the mesh shown in the figure is:}$

Options:

1. $7.2 \, \Omega$
2. $16 \, \Omega$
3. $30 \, \Omega$
4. $4.8 \, \Omega$

Solution:

$\text{Hint: Identify the series and parallel combinations of the resistors.}$ $\text{Step: Find the net resistance of the circuit.}$ $\text{On simplifying the circuit we get:}$ $\text{When } 4 \, \Omega \text{ and } 8 \, \Omega \text{ are in series we get total resistance } = 12 \, \Omega$ $\text{Now, the } 12 \, \Omega \text{ resistance is parallel to } 6 \, \Omega , \text{ net resistance:}$ $\Rightarrow \frac{1}{6} + \frac{1}{12} \Rightarrow 4 \, \Omega$ $\text{All the resistances are in series so the net resistance of the circuit is given by:}$ $R_{\text{net}} = R_1 + R_2 + R_3$ $\Rightarrow R_{\text{net}} = 4 + 4 + 8$ $\Rightarrow R_{\text{net}} = 16 \, \Omega$ $\text{Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}