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Current Question (ID: 20643)

Question:

$\text{An electromagnetic wave of frequency } 3 \text{ GHz enters a dielectric medium of relative electric permittivity } 2.25 \text{ from vacuum. The wavelength of this wave in that medium will be:}$

Options:

1. $3.45 \text{ m}$
2. $3.45 \text{ cm}$
3. $6.67 \text{ m}$
4. $6.67 \text{ cm}$

Solution:

$\text{Hint: Frequency remains the same in both mediums.}$ $\ell \text{ in vacuum } = \frac{c}{f} = \frac{3 \times 10^8}{3 \times 10^9} = 0.1 \text{ m}$ $\therefore \lambda \text{ in medium } = \frac{0.1}{\mu}$ $\text{Where refractive index } \mu = \sqrt{\mu_r \varepsilon_r}$ $\text{Assuming non-magnetic material } \mu_r = 1$ $\therefore \mu = \sqrt{2.25} = 1.5$ $\lambda_m = \frac{0.1}{1.5} = \frac{1}{15} \text{ m} = 6.67 \text{ cm}$ $= 667 \times 10^{-2} \text{ cm}$

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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
}