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Current Question (ID: 19969)

Question:
$\text{A string of length } 2.0 \text{ m, fixed at both ends is driven by a } 240 \text{ Hz vibrator.}$ $\text{If the string vibrates in its third harmonic mode, the speed of the wave and its fundamental frequency, respectively, are:}$
Options:
  • 1. $320 \text{ m/s, } 80 \text{ Hz}$
  • 2. $180 \text{ m/s, } 120 \text{ Hz}$
  • 3. $320 \text{ m/s, } 120 \text{ Hz}$
  • 4. $180 \text{ m/s, } 80 \text{ Hz}$
Solution:
$\text{Hint: } f_n = n \left( \frac{v}{2L} \right)$ $\text{Step 1: Find the speed of the wave.}$ $\text{The third harmonic mode in the string is shown in the figure below;}$ $\frac{\lambda}{2} + \frac{\lambda}{2} + \frac{\lambda}{2} = L$ $\Rightarrow \frac{3\lambda}{2} = L$ $\Rightarrow \lambda = \frac{2L}{3} = \frac{4}{3} \text{ m}$ $\text{The speed of the wave is given by;}$ $v = \lambda f$ $\Rightarrow v = \frac{4}{3} \times 240 = 320 \text{ m/s}$ $\text{Step 2: Find the fundamental frequency of the wave.}$ $\text{The fundamental frequency of the wave on the string is given by;}$ $f_0 = \frac{v}{2L}$ $3f_0 = 240 \text{ Hz}$ $\Rightarrow f_0 = \frac{240}{3} = 80 \text{ Hz}$ $\text{Hence, option (1) is the correct answer.}$

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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
}