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Current Question (ID: 16984)

Question:

$\text{A conductor } ABOCD \text{ moves along its bisector with a velocity of } 1 \text{ m/s through a perpendicular magnetic field of } 1 \text{ wb/m}^2, \text{ as shown in fig. If all the four sides are of } 1 \text{ m length each, then the induced emf between points } A \text{ and } D \text{ is:}$

Options:

1. $0$
2. $1.41 \text{ volt}$
3. $0.71 \text{ volt}$
4. $\text{None of the above}$

Solution:

$\text{There is no induced emf in the part } AB \text{ and } CD \text{ because they are moving along their length while emf induced between } B \text{ and } C \text{ i.e., between } A \text{ and } D \text{ can be calculated as follows:}$ $\text{Induced emf between } B \text{ and } C = \text{Induced emf between } A \text{ and } B$ $= Bv \left( \sqrt{2} l \right) = 1 \times 1 \times 1 \times \sqrt{2} = 1.41 \text{ volt.}$

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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
}