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Current Question (ID: 16012)

Question:

$\text{A vibrating tuning fork of frequency } n \text{ is placed near the open end of a long cylindrical tube.}$ $\text{The tube has a side opening and is also fitted with a movable reflecting piston.}$ $\text{As the piston is moved through } 8.75 \text{ cm, the intensity of sound changes from}$ $\text{a maximum to a minimum. If the speed of sound is } 350 \text{ metre per second,}$ $\text{then } n \text{ is:}$

Options:

1. $500 \text{ Hz}$
2. $1000 \text{ Hz}$
3. $2000 \text{ Hz}$
4. $4000 \text{ Hz}$

Solution:

$\text{Hint: } \Delta \phi = \frac{2\pi}{\lambda} (\Delta x)$ $\text{Step: Find the frequency of the tuning fork.}$ $\text{When the piston is moved through a distance of } 8.75 \text{ cm, the path}$ $\text{difference produced is } 2 \times 8.75 \text{ cm } = 17.5 \text{ cm.}$ $\text{Therefore, this path difference produced must be equal to } \frac{\lambda}{2} \text{ for a change in}$ $\text{the intensity of sound changes from a maximum to a minimum.}$ $\Rightarrow \frac{\lambda}{2} = \Delta x \Rightarrow \lambda = 2 \times 17.5 = 35 \text{ cm.}$ $\text{The frequency of the tuning fork is given by;}$ $\nu = \frac{v}{\lambda} = \frac{350}{.35} = 1000 \text{ Hz}$ $\text{Therefore, the frequency of the tuning fork is } 1000 \text{ Hz.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}