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Current Question (ID: 15997)

Question:

$\text{A tuning fork with a frequency of } 800 \text{ Hz produces resonance in a resonance column tube with the upper end open and the lower end closed by the water surface. Successive resonances are observed at lengths of } 9.75 \text{ cm, } 31.25 \text{ cm, and } 52.75 \text{ cm. The speed of the sound in the air is:}$

Options:

1. $500 \text{ m/s}$
2. $156 \text{ m/s}$
3. $344 \text{ m/s}$
4. $172 \text{ m/s}$

Solution:

$\text{Hint: } f = \frac{(2n + 1) \frac{v}{2(l+e)}}{}$ $\text{Step: Find the speed of the sound in the air.}$ $\text{The frequency of the tuning fork is given by: } f = \frac{(2n + 1) \frac{v}{2(l+e)}}{}$ $l_1 + e = (2n + 1) \frac{\lambda}{4} \quad \cdots (1)$ $l_2 + e = (2n + 3) \frac{\lambda}{4} \quad \cdots (2)$ $\text{From the equations (1) and (2) we get;}$ $l_2 - l_1 = \frac{\lambda}{2} = \frac{v}{2f}$ $\Rightarrow v = 2f(l_2 - l_1) = 2 \times 800 \times (31.25 - 9.75) \times 10^{-2}$ $\Rightarrow v = 344 \text{ m/s}$ $\text{Therefore, the speed of the sound in the air is } 344 \text{ m/s.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}