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Current Question (ID: 17279)

Question:
$\text{An alternating current is given as } i = i_1 \cos \omega t - i_2 \sin \omega t. \text{ The value of rms current is given by:}$
Options:
  • 1. $\frac{1}{\sqrt{2}} (i_1 + i_2)$
  • 2. $\frac{1}{\sqrt{2}} (i_1 + i_2)^2$
  • 3. $\frac{1}{\sqrt{2}} (i_1^2 + i_2^2)^{1/2}$
  • 4. $\frac{1}{2} (i_1^2 + i_2^2)^{1/2}$
Solution:
$i = i_1 \cos \omega t - i_2 \cos \left( \omega t - \frac{\pi}{2} \right)$ $\text{Here, angle between } i_1 \text{ and } i_2 = \frac{\pi}{2}$ $i_{\text{net}} = \sqrt{i_1^2 + i_2^2 + 2i_1i_2 \cos \frac{\pi}{2}} = \sqrt{i_1^2 + i_2^2}$ $i_{\text{rms}} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$

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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
}