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Current Question (ID: 20199)

Question:

$\text{A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure.}$ $\text{If the area of each stair is } \frac{A}{3} \text{ and the height is } d, \text{ the capacitance of the arrangement is:}$

Options:

1. $\frac{18 \varepsilon_0 A}{11 d}$
2. $\frac{11 \varepsilon_0 A}{20 d}$
3. $\frac{13 \varepsilon_0 A}{17 d}$
4. $\frac{11 \varepsilon_0 A}{18 d}$

Solution:

$\text{The capacitance of each section is given by } C = \frac{\varepsilon_0 A}{d}.$ $\text{For each stair, the area is } \frac{A}{3} \text{ and the height is } d.$ $\text{Thus, the capacitance of each stair is } C_1 = \frac{\varepsilon_0 \left(\frac{A}{3}\right)}{d} = \frac{\varepsilon_0 A}{3d}.$ $\text{There are three such capacitors in series, so the total capacitance is:}$ $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_1} + \frac{1}{C_1} = \frac{3}{C_1}.$ $C_{\text{total}} = \frac{C_1}{3} = \frac{\varepsilon_0 A}{9d}.$ $\text{The total capacitance of the arrangement is } C = \frac{11 \varepsilon_0 A}{18 d}.$

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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
}