Import Question JSON

Current Question (ID: 16023)

Question:

$\text{Two sitar strings, } A \text{ and } B, \text{ playing the note Ga, are slightly out of tune and produce } 6 \text{ Hz beats. The tension in the string } A \text{ is slightly reduced, and the beat frequency is found to be reduced to } 3 \text{ Hz. If the original frequency of } A \text{ is } 324 \text{ Hz, what is the frequency of } B?$

Options:

1. $316 \text{ Hz}$
2. $318 \text{ Hz}$
3. $319 \text{ Hz}$
4. $314 \text{ Hz}$

Solution:

$\text{Hint: Beat frequency is the absolute value of the difference in frequencies of the strings.}$ $\text{Step: Find the variation of frequency as a function of tension in the string and find the frequency of } B \text{ string.}$ $\text{The frequency of the wave is given as:}$ $\Rightarrow f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$ $\text{As, } f_A - f_B = 6$ $\Rightarrow f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$ $\Rightarrow f \propto \sqrt{T}$ $\text{As the tension in the string reduces, the frequency will also reduce.}$ $\text{The frequency of the } B \text{ string is given as:}$ $\Rightarrow f_A - f_B = 6$ $\Rightarrow f_B = 318 \text{ Hz}$ $\text{Therefore, the frequency of the string } B \text{ is } 318 \text{ Hz.}$ $\text{Hence, option (2) is the correct answer.}$

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
}