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Current Question (ID: 17013)

Question:

$\text{A coil is wound of a frame of rectangular cross-section. If the linear dimensions of the frame are doubled and the number of turns per unit length of the coil remains the same, then the self inductance increases by a factor of:}$

Options:

1. $6$
2. $12$
3. $8$
4. $16$

Solution:

$\text{The self inductance } L \text{ of a coil is given by } L = \mu n^2 A l, \text{ where } \mu \text{ is the permeability, } n \text{ is the number of turns per unit length, } A \text{ is the area of cross-section, and } l \text{ is the length of the coil.}$ $\text{If the linear dimensions are doubled, the area } A \text{ becomes } 4A \text{ and the length } l \text{ becomes } 2l. \text{ Thus, } L' = \mu n^2 (4A) (2l) = 8 \mu n^2 A l = 8L. \text{ Therefore, the self inductance increases by a factor of } 8.$

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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
}