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Current Question (ID: 16579)

Question:

$\text{The capacitance of a parallel plate capacitor with air as a medium is } 6 \, \mu\text{F.}$ $\text{With the introduction of a dielectric medium, the capacitance becomes } 30 \, \mu\text{F.}$ $\text{The permittivity of the medium is:}$ $\left( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2} \right)$

Options:

1. $1.77 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$
2. $0.44 \times 10^{-10} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$
3. $5.00 \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$
4. $0.44 \times 10^{-13} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$

Solution:

$\text{Hint: } C = \frac{\varepsilon_0 A}{d}$ $\text{Step: Find the permittivity of the medium.}$ $\text{The capacitance of the capacitor is written as;}$ $C = \frac{\varepsilon_0 A}{d}$ $\text{The capacitance of a parallel plate capacitor with air as a medium is } 6 \, \mu\text{F.}$ $\frac{C_{\text{air}}}{C_{\text{med}}} = \frac{A\varepsilon_0/d}{A\varepsilon/d} = \frac{\varepsilon_0}{\varepsilon}$ $\Rightarrow \frac{6}{30} = \frac{8.85 \times 10^{-12}}{\varepsilon}$ $\Rightarrow \varepsilon = 0.44 \times 10^{-10} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$ $\text{Hence, option (2) is the correct answer.}$

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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
}