Import Question JSON

Current Question (ID: 20152)

Question:

$\text{A metallic slab of thickness } \frac{2d}{3} \text{ and area of surface same as that of plates of capacitor of capacitance } C_1 \text{ is inserted parallel to plates of capacitor such that its new capacitance becomes equal to } C_2. \text{ If } d \text{ is space width between the two plates, then } \frac{C_2}{C_1} \text{ is equal to:}$ $1. \text{ 1}$ $2. \text{ 2}$ $3. \text{ 3}$ $4. \text{ 4}$

Options:

1. $1$
2. $2$
3. $3$
4. $4$

Solution:

$\text{Hint: The capacitance of the parallel plate capacitor is given as } C = \frac{\epsilon_0 A}{d}. $ $\text{Explanation: For the parallel plate capacitor of separation between plates } d \text{ and the area between the plates is } A. \text{ The capacitance of the capacitor is given as } C_1 = \frac{\epsilon_0 A}{d}. $ $\text{Now metallic slab of thickness } \frac{2d}{3} \text{ is inserted in a parallel plate capacitor. Now thickness is given as } d_{\text{eff}} = d - \frac{2d}{3} = \frac{d}{3}. $ $\text{The capacitance of the new configuration of parallel plate capacitor is given as } C_2 = \frac{\epsilon_0 A}{d_{\text{eff}}} = \frac{\epsilon_0 A}{d/3} = \frac{3 \epsilon_0 A}{d}. $ $\text{The ratio of capacitance is } \frac{C_2}{C_1} = 3 $ $\text{Option (3) is correct.}$

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
}