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Current Question (ID: 16573)

Question:

$\text{A parallel plate capacitor with cross-sectional area } A \text{ and separation } d \text{ has air between the plates. An insulating slab of the same area but the thickness of } \frac{d}{2} \text{ is inserted between the plates as shown in the figure, having a dielectric constant, } K = 4. \text{ The ratio of the new capacitance to its original capacitance will be:}$

Options:

1. $2 : 1$
2. $8 : 5$
3. $6 : 5$
4. $4 : 1$

Solution:

$\text{Hint: } C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{k}}$ $\text{Step: Find the new capacitance of the capacitor.}$ $\text{The capacitance is given by: } C = \frac{\varepsilon_0 A}{d}$ $\text{The original capacitance of the capacitor is: } C_i = \frac{\varepsilon_0 A}{d} \quad \cdots (1)$ $\text{The new capacitance of the capacitor will be: } C_f = \frac{\varepsilon_0 A}{d - t + \frac{t}{k}}$ $\text{Substituting the values, we get:}$ $\Rightarrow C_f = \frac{\varepsilon_0 A}{d - \frac{d}{2} + \frac{d}{8}}$ $\Rightarrow C_f = \frac{8 \varepsilon_0 A}{5d} \quad \cdots (2)$ $\text{Comparing the values of equations (1) and (2), we get:}$ $\Rightarrow C_f = \frac{8 C_i}{5} \Rightarrow \frac{C_f}{C_i} = \frac{8}{5}$ $\text{Hence, option (2) is the correct answer.}$

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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
}