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Current Question (ID: 16652)

Question:
$\text{In the figure, the value of resistors to be connected between } C \text{ and } D \text{ so that}$ $\text{the resistance of the entire circuit between } A \text{ and } B \text{ does not change with the}$ $\text{number of elementary sets used is:}$
Options:
  • 1. $R$
  • 2. $R(\sqrt{3} - 1)$
  • 3. $3R$
  • 4. $R(\sqrt{3} + 1)$
Solution:
$\text{Cut the series from XY and let the resistance towards right of XY be } R_0$ $\text{whose value should be such that when connected across } AB \text{ does not}$ $\text{change the entire resistance. The combination is reduced to as shown}$ $\text{below.}$ $R_0 \text{ and } 2R \text{ are in series and their sum is in parallel with } R.$ $\text{Thus, } R_{AB} = \frac{(R_0 + 2R)R}{R_0 + 2R + R} = \frac{R_0 R + 2R^2}{R_0 + 3R} = R_0$ $\Rightarrow R_0^2 + 2RR_0 - 2R^2 = 0 \Rightarrow R_0 = R(\sqrt{3} - 1)$

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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
}