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Current Question (ID: 20281)

Question:

$\text{The drift speed of electrons, when } 1.5 \text{ A of current flows in a copper wire of cross-section } 5 \text{ mm}^2, \text{ is } v. \text{ If the electron density in copper is } 9 \times 10^{28}/\text{m}^3, \text{ then the value of drift speed } v \text{ in mm/s is close to: (take charge of electron to be } e = 1.6 \times 10^{-19} \text{ C)}$

Options:

1. $0.02$
2. $3$
3. $2$
4. $0.2$

Solution:

$\text{Hint: } I = neAv_d$ $I = 1.5 \text{ A}$ $A = 5 \times 10^{-6} \text{ m}^2$ $n = 9 \times 10^{28} \text{ m}^{-3}$ $e = 1.6 \times 10^{-19} \text{ C}$ $v_d = \frac{I}{neA} = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}} \approx 0.02 \text{ mm/s}$

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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
}