Import Question JSON

Current Question (ID: 17288)

Question:

$\text{Match List I (expression for current) with List II (rms value of current) and select the correct answer.}$ $\text{List I}$ \begin{array}{ll} (a) & I = I_0 \sin \omega t \cos \omega t \\ (b) & I = I_0 \sin \left( \omega t + \frac{\pi}{3} \right) \\ (c) & I_0 (\sin \omega t + \cos \omega t) \\ (d) & I = I_0 (e) \end{array} \quad \text{List II} \begin{array}{ll} (i) & I_0 \\ (ii) & \frac{I_0}{\sqrt{2}} \\ (iii) & I_0 e \\ (iv) & \frac{I_0}{2\sqrt{2}} \end{array}$

Options:

1. $(iv) \ (ii) \ (i) \ (iii)$
2. $(iv) \ (i) \ (iii) \ (i)$
3. $(ii) \ (iv) \ (iii) \ (i)$
4. $(ii) \ (iv) \ (i) \ (iii)$

Solution:

$(A) \ I = I_0 \sin \omega t \cdot \cos \omega t = \frac{I_0}{2} \cdot 2 \sin \omega t \cdot \cos \omega t = \frac{I_0}{2} \sin 2 \omega t$ $I_{\text{rms}} = \frac{I_0/2}{\sqrt{2}} = \frac{I_0}{2\sqrt{2}}$ $(B) \ I = I_0 \sin \left( \omega t + \frac{\pi}{3} \right)$ $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ $(C) \ I = I_0 (\sin \omega t + \cos \omega t) = I_0 \sin \omega t + I_0 \cos \omega t$ $\text{Multiplied and divided by } 2 \sqrt{I_0^2}$ $I = \sqrt{2I_0^2} \left[ \frac{I_0}{\sqrt{2I_0^2}} \sin \omega t + \frac{I_0}{\sqrt{2I_0^2}} \cos \omega t \right]$ $I = I_0 \sqrt{2} \sin(\omega t + \alpha)$ $I_{\text{rms}} = \frac{I_0 \sqrt{2}}{\sqrt{2}} = I_0$ $(D) \ I = I_0 (e)$ $I_{\text{rms}} = I_0$

Import JSON File

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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
}