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Current Question (ID: 16006)

Question:
$\text{A pipe closed at one end produces a fundamental note of } 412 \text{ Hz. It is cut into two pieces of equal length. The fundamental nodes produced by the two pieces are:}$ $1.\ 206 \text{ Hz, } 412 \text{ Hz}$ $2.\ 412 \text{ Hz, } 824 \text{ Hz}$ $3.\ 206 \text{ Hz, } 824 \text{ Hz}$ $4.\ 824 \text{ Hz, } 1648 \text{ Hz}$
Options:
  • 1. $206 \text{ Hz, } 412 \text{ Hz}$
  • 2. $412 \text{ Hz, } 824 \text{ Hz}$
  • 3. $206 \text{ Hz, } 824 \text{ Hz}$
  • 4. $824 \text{ Hz, } 1648 \text{ Hz}$
Solution:
$\text{Hint: } f_{\text{closed}} = \frac{v}{4l} \text{ and } f_{\text{open}} = \frac{v}{2l}$ $\text{Step 1: Find the fundamental frequency of the original pipe.}$ $f_0 = \frac{v}{4l} = 412 \text{ Hz}$ $\text{Step 2: Find the fundamental frequency for new pipes.}$ $f'_{\text{open}} = \frac{v}{2\left(\frac{l}{2}\right)} = 1648 \text{ Hz}$ $f'_{\text{closed}} = \frac{v_0}{4\left(\frac{l}{2}\right)} = \frac{v_0}{2l} = 824 \text{ Hz}$

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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
}