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Current Question (ID: 15994)

Question:

$\text{The fundamental frequency of a closed organ pipe of a length } 20 \text{ cm is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:}$

Options:

1. $80 \text{ cm}$
2. $100 \text{ cm}$
3. $120 \text{ cm}$
4. $140 \text{ cm}$

Solution:

$\text{Hint: } \nu_C = \frac{v}{4l}$ $\text{Step: Find the length of the organ pipe open at both ends.}$ $\text{The fundamental frequencies of closed and open pipes are given as:}$ $\nu_C = \frac{v}{4l}$ $\nu_O = \frac{v}{2l'}$ $\text{Given the second overtone, (i.e. third harmonic) of the open pipe is equal to the fundamental frequency of the closed pipe}$ $\text{ie., } 3\nu_O = \nu_C$ $\Rightarrow 3 \frac{v}{2l'} = \frac{v}{4l}$ $\Rightarrow l' = 6l = 6 \times 20 = 120 \text{ cm}$ $\text{Therefore, the length of the organ pipe open at both ends is } 120 \text{ cm.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}