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Current Question (ID: 16541)

Question:

$\text{In the given figure if } V = 4 \text{ volt each plate of the capacitor has a surface area of } 10^{-2} \text{ m}^2 \text{ and the plates are } 0.1 \times 10^{-3} \text{ m apart, then the number of excess electrons on the negative plate is:}$

Options:

1. $5.15 \times 10^9$
2. $2.21 \times 10^{10}$
3. $3.33 \times 10^9$
4. $2.21 \times 10^9$

Solution:

$\text{The capacitance } C \text{ is given by } C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 10^{-2}}{0.1 \times 10^{-3}} = 8.85 \times 10^{-12} \text{ F}$ $\text{The charge } Q \text{ is given by } Q = CV = 8.85 \times 10^{-12} \times 4 = 3.54 \times 10^{-11} \text{ C}$ $\text{The number of electrons } n \text{ is given by } n = \frac{Q}{e} = \frac{3.54 \times 10^{-11}}{1.6 \times 10^{-19}} = 2.21 \times 10^{10}$

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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
}