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Current Question (ID: 20520)

Question:

$\text{Two coils of self-inductance } L_1 \text{ and } L_2 \text{ are connected in series}$ $\text{combination having a mutual inductance of the coils as } M. \text{ The}$ $\text{equivalent self-inductance of the combination will be:}$

Options:

1. $\frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{M}$
2. $L_1 + L_2 + M$
3. $L_1 + L_2 + 2M$
4. $L_1 + L_2 - 2M$

Solution:

$\text{For series combination of inductors with mutual inductance,}$ $\text{the equivalent inductance is given by:}$ $L = L_1 + L_2 - 2M$

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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
}