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Current Question (ID: 16007)

Question:

$\text{The two nearest harmonics of a tube close at one end and open at the other end are } 220 \text{ Hz and } 260 \text{ Hz. What is the fundamental frequency of the system?}$

Options:

1. $10 \text{ Hz}$
2. $20 \text{ Hz}$
3. $30 \text{ Hz}$
4. $40 \text{ Hz}$

Solution:

$\text{Hint: } f = \left(2n + 1\right) \frac{v}{4l}$ $\text{Step: Find the fundamental frequency of the system.}$ $\text{Frequency of nth overtone in closed-end tube}$ $\Rightarrow f = \frac{\left(2n+1\right) v}{4l} \quad (\text{where, } n = 1, 2, 3, \ldots)$ $\text{Now, given two nearest harmonics are of frequency } 220 \text{Hz and } 260 \text{Hz.}$ $\text{So, } \frac{\left(2n+1\right) v}{4l} = 220 \text{Hz} \quad \ldots (i)$ $\text{Next harmonics occur at,}$ $\frac{\left(2n+3\right) v}{4l} = 260 \text{Hz} \quad \ldots (ii)$ $\text{On subtracting Eq. (i) from Eq. (ii), we get:}$ $\frac{\{ \left(2n+3\right) - \left(2n+1\right) \} v}{4l} = 260 - 220$ $2 \left(\frac{v}{4l}\right) = 40 \Rightarrow \frac{v}{4l} = 20 \text{Hz}$ $\text{Therefore, the fundamental frequency of the system is } 20 \text{ Hz.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}