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Current Question (ID: 20616)

Question:

$\text{An electromagnetic wave of frequency } 5 \text{ GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are } 2. \text{ Its velocity in this medium is:}$ $1. \ 15 \times 10^7 \ \text{m/s}$ $2. \ 5 \times 10^7 \ \text{m/s}$ $3. \ 5 \times 10^8 \ \text{m/s}$ $4. \ 15 \times 10^8 \ \text{m/s}$

Options:

1. $15 \times 10^7 \ \text{m/s}$
2. $5 \times 10^7 \ \text{m/s}$
3. $5 \times 10^8 \ \text{m/s}$
4. $15 \times 10^8 \ \text{m/s}$

Solution:

$\text{Hint: } v = \frac{1}{\sqrt{\mu_r \epsilon_r}} = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$ $\text{Given: Frequency of wave } f = 5 \text{ GHz}$ $= 5 \times 10^9 \ \text{Hz}$ $\text{Relative permittivity, } \epsilon_r = 2$ $\text{and Relative permeability, } \mu_r = 2$ $\text{Since speed of light in a medium is given by,}$ $V = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_r \mu_0 \epsilon_r \epsilon_0}}$ $v = \frac{1}{\sqrt{\mu_r \epsilon_r}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$ $\text{Where } C \text{ is speed of light in vacuum.}$ $\therefore v = \frac{3 \times 10^8}{\sqrt{4}} = \frac{30 \times 10^7}{2} \ \text{m/s}$ $= 15 \times 10^7 \ \text{m/s}$ $\therefore \text{ Ans. is } 15$

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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
}