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Current Question (ID: 20540)

Question:

$\text{An inductance coil has a reactance of } 100 \, \Omega. \text{ When an AC signal of frequency } 1000 \, \text{Hz is applied to the coil, the applied voltage leads the current by } 45^\circ. \text{ The self-inductance of the coil is:}$

Options:

1. $1.1 \times 10^{-2} \, \text{H}$
2. $1.1 \times 10^{-1} \, \text{H}$
3. $5.5 \times 10^{-5} \, \text{H}$
4. $6.7 \times 10^{-7} \, \text{H}$

Solution:

$\text{Hint: } \tan \theta = \frac{X_L - X_C}{R}$ $\text{Reactance of inductance coil } = \sqrt{R^2 + x_L^2} = 100$ $f = 1000 \, \text{Hz of applied AC signal}$ $\text{Voltage leads current by } 45^\circ$ $\tan 45^\circ = \frac{i X_L}{i R} = \frac{\omega L}{R}$ $R = X_L = \omega L$ $\sqrt{2} X_L = 100 \Rightarrow X_L = 50 \sqrt{2}$ $L = \frac{50 \sqrt{2}}{\omega} = \frac{50 \sqrt{2}}{2 \pi f} = \frac{25 \sqrt{2}}{\pi \times 1000} \, \text{H}$ $= 1.125 \times 10^{-2} \, \text{H}$

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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
}