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Current Question (ID: 19966)

Question:

$\text{A granite rod of } 60 \text{ cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is } 2.7 \times 10^3 \text{ kg/m}^3 \text{ and it's Young's modulus is } 9.27 \times 10^{10} \text{ Pa.}$ $\text{What will be the fundamental frequency of the longitudinal vibrations?}$

Options:

1. $5 \text{ kHz}$
2. $2.5 \text{ kHz}$
3. $10 \text{ kHz}$
4. $7.5 \text{ kHz}$

Solution:

$\text{Hint: } v = n \lambda$ $\frac{\lambda}{4} = 0.3$ $\lambda = 1.2 \text{ m}$ $v = n \lambda$ $n = \frac{v}{\lambda} = \frac{1}{1.2} \sqrt{\frac{Y}{d}}$ $= \frac{1}{1.2} \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^3}} = 4.8 \text{ kHz} \approx 5 \text{ kHz}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}