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Current Question (ID: 16025)

Question:

$\text{Two organ pipes closed at one end produce } 5 \text{ beats per second in fundamental mode. If the ratio of their lengths is } 10 : 11, \text{ then their frequencies (in Hz) are:}$

Options:

1. $55, 50$
2. $105, 100$
3. $75, 70$
4. $100, 95$

Solution:

$\text{Hint: Beat frequency per second} = |f_1 - f_2|$ $\text{Step 1: Relate the length of two pipes.}$ $\frac{l_1}{l_2} = \frac{10}{11}$ $\Rightarrow l_2 = \frac{11}{10} l_1$ $\text{Step 2: Find the fundamental frequency for both pipes.}$ $f_1 = \frac{v_0}{4l_1} \text{ and } f_2 = \frac{v_0}{4l_2}$ $\text{Step 3: Find the beat frequency}$ $\Delta f = f_1 - f_2$ $5 = \frac{v_0}{4l_1} - \frac{v_0}{4} \left( \frac{10}{11} \frac{1}{l_1} \right)$ $5 = \frac{v_0}{4l_1} \left( \frac{1}{11} \right)$ $\Rightarrow \frac{v_0}{l_1} = 220$ $\text{Step 4: Find the fundamental frequencies.}$ $f_1 = \frac{v_0}{4l_1} = 55 \text{ Hz}$ $f_2 = \frac{v_0}{4l_1} \left( \frac{10}{11} \right)$ $= 55 \times \frac{10}{11} = 50 \text{ Hz}$ $\text{Therefore, the fundamental frequencies of the two waves are } 55 \text{ Hz and } 50 \text{ Hz respectively.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}