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Current Question (ID: 16004)

Question:

$A \ 1 \ \text{m, long tube open at one end with a movable piston at the other end}$ $\text{shows resonance with a fixed frequency source (a tuning fork of frequency}$ $340 \ \text{Hz}) \ \text{when the minimum tube length is} \ 25.5 \ \text{cm. The speed of sound in air}$ $\text{at the temperature of the experiment is:}$ $(\text{The edge effects may be neglected.})$

Options:

1. $324.16 \ \text{m/s}$
2. $320 \ \text{m/s}$
3. $345 \ \text{m/s}$
4. $346.8 \ \text{m/s}$

Solution:

$\text{Hint: Displacement node is formed at the closed end whereas antinode is}$ $\text{formed at the open end.}$ $\text{Step 1: Find the wavelength of the sound wave.}$ $\text{Such a system produces odd harmonics. The fundamental node in the}$ $\text{closed pipe is given by;} \ L = \frac{\lambda}{4}$ $\text{Given that the minimum length of the tube is} \ L = 25.5 \ \text{cm} = 0.255 \ \text{m, we}$ $\text{can find the wavelength:}$ $\lambda = 4L = 4 \times 0.255 \ \text{m} = 1.02 \ \text{m}$ $\text{Step 2: Find the speed of the sound wave.}$ $\text{The speed of sound} \ v \ \text{can be calculated using the formula:} \ v = f \cdot \lambda$ $\text{Given that a tuning fork of frequency} \ 340 \ \text{Hz.}$ $\text{Substituting the known values into the equation:} \ v = f \cdot \lambda$ $v = 340 \times 1.02 = 346.8 \ \text{m/s}$ $\text{Hence, option (4) is the correct answer.}$

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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
}