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Current Question (ID: 20558)

Question:

$\text{The variation of impedance } (Z) \text{ with angular frequency } (\omega) \text{ for two electrical elements is shown in the graph given.}$ $\text{If } X_L, X_C \text{ and } R \text{ are inductive reactance, capacitive reactance, and resistance respectively then:}$

Options:

1. $A \text{ is resistor, } B \text{ is inductor}$
2. $A \text{ is inductor, } B \text{ is capacitor}$
3. $A \text{ is inductor, } B \text{ is resistor}$
4. $A \text{ is capacitor, } B \text{ is inductor}$

Solution:

$\text{Hint: } X_C = \frac{1}{\omega C}$ $\text{Step: Find the elements based on their characteristics.}$ $\text{We need to match the variation of impedance } (Z) \text{ with angular frequency } (\omega) \text{ for two elements based on their characteristics.}$ $\text{For a resistor } (R): \text{ The impedance } (Z) \text{ of a resistor is constant and independent of frequency.}$ $\text{So, if } (Z) \text{ is a straight horizontal line on the graph, it represents a resistor.}$ $\text{For an inductor } (L): \text{ The inductive reactance is } X_L = \omega L, \text{ meaning the impedance increases linearly with increasing frequency.}$ $\text{For a capacitor } (C): \text{ The capacitive reactance is } X_C = \frac{1}{\omega C}, \text{ meaning the impedance decreases as frequency increases.}$ $\text{Thus, } A \text{ is an inductor and } B \text{ is a capacitor.}$ $\text{Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}