Import Question JSON

Current Question (ID: 16002)

Question:
$\text{A cylindrical tube open at both ends has a fundamental frequency } f_0 \text{ in the air.}$ $\text{The tube is dipped vertically in water such that half its length is inside water.}$ $\text{The fundamental frequency of the air column now will be:}$
Options:
  • 1. $\frac{3f_0}{4}$
  • 2. $f_0$
  • 3. $\frac{f_0}{2}$
  • 4. $2f_0$
Solution:
$\text{Hint: Recall the concept of an organ pipe.}$ $\text{Step 1: Write the fundamental frequencies in the case of a closed and open organ pipe.}$ $\text{For closed organ pipe and open organ pipe, fundamental frequencies are}$ $f_0 = \frac{v}{2l} \text{ and } f_0 = \frac{v}{4l} \text{ respectively.}$ $\text{Open pipe:}$ $f_0 = \frac{v}{2l}$ $\text{Closed pipe:}$ $\text{Step 2: Find the length with the help of a diagram for the closed organ pipe.}$ $\text{Here as the length is half so,}$ $f'_0 = \frac{v}{4\left(\frac{l}{2}\right)} = \frac{v}{2l}$ $\text{Step 3: If the tube is dipped vertically in the water then,}$ $\text{The fundamental frequency of the air column } = f_0$ $\text{Hence, option (2) is the correct answer.}$

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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
}