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Current Question (ID: 17029)

Question:
$\text{A small square loop of wire of side } l \text{ is placed inside a large square loop of side } L \ (L \gg l).$ $\text{If the loops are coplanar and their centres coincide, the mutual inductance of the system is directly proportional to:}$
Options:
  • 1. $\frac{L}{l}$
  • 2. $\frac{l}{L}$
  • 3. $\frac{L^2}{l}$
  • 4. $\frac{l^2}{L}$
Solution:
$\text{Step 1: Use analogy between circular \& square coil}$ $B_C = \frac{\mu_0 i}{2r}$ $B \propto \frac{1}{L}$ $B_C = \frac{kI}{L}$ $\phi_{21} = B_C A$ $= \frac{k l^2}{L} I$ $\text{Step 2: Use formula for mutual inductance}$ $\phi_{21} = M I_1$ $M \propto \frac{l^2}{L}$

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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
}