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Current Question (ID: 19980)

Question:

$\text{An observer is riding on a bicycle and moving towards a hill at } 18 \text{ kmh}^{-1}. \text{ He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by the source is } 640 \text{ Hz and the velocity of the sound in air is } 320 \text{ m/s, the beat frequency between the two sounds heard by the observer will be:}$

Options:

1. $20 \text{ Hz}$
2. $40 \text{ Hz}$
3. $60 \text{ Hz}$
4. $10 \text{ Hz}$

Solution:

$\text{Hint: Use Doppler effect formula.}$ $\text{Step 1: Find the direct frequency heard by the observer.}$ $f_1 = f_0 \left(\frac{320 - 5}{320}\right) = 630 \text{ Hz}$ $\text{Step 2: Find the reflected frequency heard by the observer.}$ $f_2 = f_0 \left(\frac{320 + 5}{320}\right) = 650 \text{ Hz}$ $\text{Step 3: Find the beat frequency.}$ $f_2 - f_1 = 650 - 630 = 20 \text{ Hz}$

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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
}