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Current Question (ID: 16653)

Question:

$\text{A ring is made of a wire having a resistance of } R_0 = 12 \, \Omega. \text{ Find points } A \text{ and } B, \text{ as shown in the figure, at which a current-carrying conductor should be connected so that the resistance } R \text{ of the subcircuit between these points equals } \frac{8}{3} \, \Omega?$

Options:

1. $\frac{l_1}{l_2} = \frac{5}{8}$
2. $\frac{l_1}{l_2} = \frac{1}{3}$
3. $\frac{l_1}{l_2} = \frac{3}{8}$
4. $\frac{l_1}{l_2} = \frac{1}{2}$

Solution:

$\text{Hint: Both parts of the ring will be in a parallel combination.}$ $\text{Step 1: Find the value of the resistances of the parts.}$ $\text{When the wire is opened:}$ $R_1 + R_2 = 12 \, \Omega$ $\frac{R_1 \times R_2}{R_1 + R_2} = \frac{8}{3} \, \Omega$ $\Rightarrow R_1 R_2 = 32 \, \Omega^2$ $\text{We get, } R_1 = 4 \, \Omega \text{ and } R_2 = 8 \, \Omega$ $\text{Step 2: Find the ratio of lengths of the parts.}$ $\text{We know, } R \propto l$ $\text{Hence, } \frac{l_1}{l_2} = \frac{R_1}{R_2} = \frac{1}{2}$

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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
}