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Current Question (ID: 16967)

Question:

$\text{The graph gives the magnitude } B(t) \text{ of a uniform magnetic field that exists throughout a conducting loop, perpendicular to the plane of the loop. Rank the five regions of the graph according to the magnitude of the emf induced in the loop, greatest first:}$

Options:

1. $b > (d = e) < (a = c)$
2. $b > (d = e) > (a = c)$
3. $b < d < e < c < a$
4. $b > (a = c) > (d = e)$

Solution:

$\text{Induced emf, } e = A \frac{dB}{dt}$ $i.e., \ e \propto \frac{dB}{dt} \ (= \text{slope of } B - t \text{ graph})$ $\text{In the given graph, slope of } AB > \text{slope of } CD, \text{ slope in the } 'a' \text{ region } = \text{slope in the } 'c' \text{ region } = 0, \text{ slope in the } 'd' \text{ region } = \text{slope in the } 'e' \text{ region } \neq 0.$ $\text{That's why emf induced in region } b > (d = e) > (a = c)$

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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
}