Import Question JSON

Current Question (ID: 16033)

Question:

$\text{A tuning fork of frequency } 512 \text{ Hz makes } 4 \text{ beats/s with the vibrating strings of a piano.}$ $\text{The beat frequency decreases to } 2 \text{ beats/s when the tension in the piano strings is slightly increased.}$ $\text{The frequency of the piano string before increasing the tension was:}$

Options:

1. $510 \text{ Hz}$
2. $514 \text{ Hz}$
3. $516 \text{ Hz}$
4. $508 \text{ Hz}$

Solution:

$\text{Beat frequency per second} = |f_1 - f_2|$ $\text{Find the frequency of the piano string before increasing the tension.}$ $\text{Let the frequency of the piano is } n_p = \text{frequency of piano.}$ $(n_p \propto \sqrt{T})$ $n_f = \text{frequency of tuning fork} = 512 \text{ Hz}$ $x = \text{Beat frequency} = 4 \text{ beats/s, which is decreasing (4} \rightarrow 2) \text{ after changing the tension of piano wire.}$ $\text{Also, the tension of the piano wire is increasing so } n_p \uparrow$ $\text{Hence, } n_p \uparrow - n_f = x \downarrow \rightarrow \text{wrong}$ $n_f - n_p \uparrow = x \downarrow \rightarrow \text{correct}$ $\Rightarrow n_p = n_f - x = 512 - 4 = 508 \text{ Hz}$ $\text{Therefore, the frequency of the piano string before increasing the tension is } 508 \text{ Hz.}$ $\text{Hence, option (4) is the correct answer.}$

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
}