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Current Question (ID: 16988)

Question:

$\text{A conducting square frame of side } a \text{ and a long straight wire carrying current } i \text{ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity } v. \text{ The emf induced in the frame will be proportional to:}$

Options:

1. $\frac{1}{x^2}$
2. $\frac{1}{(2x-a)^2}$
3. $\frac{1}{(2x+a)^2}$
4. $\frac{1}{(2x-a) \times (2x+a)}$

Solution:

$\text{Potential difference across } PQ: \\ V_P - V_Q = B_1 av = \frac{\mu_0 I}{2\pi \left( x - \frac{a}{2} \right)} av$ $\text{Potential difference across } RS: \\ V_S - V_R = B_2 av = \frac{\mu_0 I}{2\pi \left( x + \frac{a}{2} \right)} av$ $V_{\text{net}} = (V_P - V_Q) - (V_S - V_R) = \frac{\mu_0 I}{2\pi} av - \frac{\mu_0 I}{2\pi} av$ $V_{\text{net}} = \frac{\mu_0 I av}{2\pi} \left( \frac{1}{x - \frac{a}{2}} - \frac{1}{x + \frac{a}{2}} \right) = \frac{\mu_0 I av}{2\pi} \left( \frac{a}{(x - \frac{a}{2})(x + \frac{a}{2})} \right)$ $V_{\text{net}} \propto \frac{1}{(2x-a) \times (2x+a)}$

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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
}