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Current Question (ID: 20002)

Question:

$\text{An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:}$ $1.\ 20\%$ $2.\ 10\%$ $3.\ 5\%$ $4.\ 0\%$

Options:

1. $20\%$
2. $10\%$
3. $5\%$
4. $0\%$

Solution:

$\text{Hint: The percentage increase in the apparent frequency is given by,}$ $\frac{\nu' - \nu}{\nu} \times 100.$ $\text{Step 1: Find the apparent frequency.}$ $f = f_0 \left( \frac{v_s + v_0}{v_s} \right)$ $f = f_0 \left[ \frac{v + \frac{v}{5}}{v} \right]$ $f = \frac{6}{5} f_0$ $\text{Step 2: Find the percentage increase in the frequency.}$ $p = \left[ \frac{\frac{6}{5} f_0 - f_0}{f_0} \right] \times 100$ $p = 20\%$

Import JSON File

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
}