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Current Question (ID: 15944)

Question:
$\text{The wave equations of two particles are given by}$ $y_1 = a \sin(\omega t - kx), \; y_2 = a \sin(kx + \omega t), \; \text{then:}$
Options:
  • 1. $\text{they are moving in the opposite direction.}$
  • 2. $\text{the phase between them is } 90^\circ.$
  • 3. $\text{the phase between them is } 45^\circ.$
  • 4. $\text{the phase between them is } 0^\circ.$
Solution:
$\text{Hint: } y = a \sin(\omega t \pm kx)$ $\text{Step: Find the phase of the wave.}$ $\text{The wave velocity of the wave is given by; } v_{p\text{max}} = A\omega$ $\text{The wave speed of the first wave is given by; } v_1 = \frac{\omega}{k} = -\frac{\omega}{k}$ $\text{The wave speed of the second wave is given by; } v_2 = \frac{\omega}{k}$ $\text{Therefore, both waves are moving opposite to each other.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}