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Current Question (ID: 16017)

Question:

$\text{The equation of a standing wave in a string is}$ $y = (200 \text{ m}) \sin\left(\frac{2\pi}{50} x\right) \cos\left(\frac{2\pi}{0.01} t\right)$ $\text{where } x \text{ is in metres and } t \text{ is in seconds.}$ $\text{At the position of antinode, how many times does the distance of a string particle become 200 m from its mean position in one second?}$

Options:

1. $100 \text{ times}$
2. $50 \text{ times}$
3. $200 \text{ times}$
4. $400 \text{ times}$

Solution:

$\text{Hint: } \omega = \frac{2\pi}{T} = 2\pi f$ $\text{Step: Find the frequency of reaching maximum displacement.}$ $\omega = \frac{2\pi}{T} = 2\pi f \Rightarrow \frac{2\pi}{0.01} = 2\pi f \Rightarrow f = 100 \text{ Hz}$ $\text{The frequency of oscillation is 100 Hz.}$ $\text{The particle will reach at maximum displacement two times in one oscillation.}$ $\text{Frequency of reaching maximum displacement} = 2 \times 100 = 200 \text{ Hz.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}