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Current Question (ID: 15993)

Question:

$\text{A string of length } 3 \text{ m and a linear mass density of } 0.0025 \text{ kg/m is fixed at both ends.}$ $\text{One of its resonance frequencies is } 252 \text{ Hz.}$ $\text{The next higher resonance frequency is } 336 \text{ Hz.}$ $\text{Then the fundamental frequency will be:}$

Options:

1. $84 \text{ Hz}$
2. $63 \text{ Hz}$
3. $126 \text{ Hz}$
4. $168 \text{ Hz}$

Solution:

$\text{Hint: } f_n = n f_1$ $\text{Step: Find the fundamental frequency of the wave.}$ $f_n = n f_1 = 252 \text{ Hz};$ $f_{n+1} = (n+1) f_1 = 336 \text{ Hz}$ $\text{Now, } \frac{f_n}{f_{n+1}} = \frac{n}{n+1} = \frac{252}{336} \Rightarrow n = 3$ $\text{Therefore, the fundamental frequency of the wave is } f_1 = \frac{252}{3} = 84 \text{ Hz.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}