Import Question JSON

Current Question (ID: 20542)

Question:

$\text{In a series } L R \text{ circuit, a power of } 400 \text{ W is dissipated from a source}$ $\text{of } 250 \text{ V and } 50 \text{ Hz. The power factor of the circuit is } 0.8. \text{ To bring}$ $\text{the power factor to unity, a capacitor of value } \left( \frac{n}{3\pi} \right) \mu \text{F is added in}$ $\text{series with the } L \text{ and } R \text{ components. The value of } n \text{ is:}$

Options:

1. $400$
2. $300$
3. $200$
4. $100$

Solution:

$\text{Hint: } P = \frac{E_{rms}^2}{Z} \cos \phi$ $\text{Step: Find the value of } n.$ $\text{The power dissipated by the circuit is given by:}$ $P = \frac{E_{rms}^2}{Z} \cos \phi$ $\Rightarrow 400 = \frac{(250)^2}{Z} \times 0.8$ $\Rightarrow Z = \frac{(250)^2}{400} \times 0.8 = \frac{250 \times 250 \times 0.8}{400}$ $\Rightarrow Z = 125 \ \Omega$ $\text{As the power factor of the circuit is unity means the circuit is in}$ $\text{resonance i.e., } X_L = X_C = Z \sin \phi$ $\Rightarrow X_L = X_C = 125 \times \frac{3}{5} \quad \left[ \cos \phi = \frac{4}{5} \right]$ $\Rightarrow X_C = 75 \ \Omega$ $\Rightarrow \frac{1}{2\pi \times 50 \times C} = 75 \ \Omega$ $\Rightarrow \frac{1}{2\pi \times 50 \times 75} = C = \frac{n}{3\pi} \times 10^{-6}$ $\Rightarrow \frac{10^6}{2 \times 50 \times 75} = \frac{n}{3}$ $\Rightarrow n = \frac{10^6}{2 \times 50 \times 25} = \frac{10000}{25} = 400$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

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
}