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Current Question (ID: 16646)

Question:

$\text{A wire of resistance } 12 \, \Omega \text{m}^{-1} \text{ is bent to form a complete circle of radius } 10 \, \text{cm.}$ $\text{The resistance between its two diametrically opposite points, } A \text{ and } B \text{ as shown in the figure, is:}$

Options:

1. $0.6\pi \, \Omega$
2. $3\pi \, \Omega$
3. $6\pi \, \Omega$
4. $6\pi \, \Omega$

Solution:

$\text{Hint: Two parts will be in the parallel combination.}$ $\text{Step 1: Find the resistance of the wire.}$ $\text{The circumference of the circle} = 2\pi r = 2\pi(0.1) = \frac{\pi}{5}$ $\text{(} R = 10\text{cm} = 0.1\text{m)}$ $\text{The resistance of the wire} = 12 \times \frac{\pi}{5} = \frac{12\pi}{5}$ $\text{Step 2: Find the resistance of each part.}$ $\text{Since the wire is divided into two sections, the resistance of each section} = \frac{12\pi}{10}$ $\text{Step 3: Find the equivalent resistance.}$ $\text{The equivalent resistance will be,}$ $R_{\text{eq}} = \frac{\frac{12\pi}{10} \times \frac{12\pi}{10}}{\frac{12\pi}{10} + \frac{12\pi}{10}} = \frac{6\pi}{10} = 0.6\pi \, \Omega$ $\text{Hence, option (1) is the correct answer.}$

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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
}