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Current Question (ID: 15934)

Question:

$\text{The equation } y(x, t) = 0.005 \cos(\alpha x - \beta t) \text{ describes a wave traveling along the } x\text{-axis. If the wavelength and the time period of the wave are } 0.08 \text{ m and } 2.0 \text{ s, respectively, then } \alpha \text{ and } \beta \text{ in appropriate units are:}$ $1. \ \alpha = 25.00\pi, \ \beta = \pi$ $2. \ \alpha = \frac{0.08}{\pi}, \ \beta = \frac{2.0}{\pi}$ $3. \ \alpha = \frac{\pi}{0.04}, \ \beta = \frac{\pi}{1.0}$ $4. \ \alpha = 12.50\pi, \ \beta = \frac{\pi}{2.0}$

Options:

1. $\alpha = 25.00\pi, \ \beta = \pi$
2. $\alpha = \frac{0.08}{\pi}, \ \beta = \frac{2.0}{\pi}$
3. $\alpha = \frac{\pi}{0.04}, \ \beta = \frac{\pi}{1.0}$
4. $\alpha = 12.50\pi, \ \beta = \frac{\pi}{2.0}$

Solution:

$\text{The standard wave equation is given as: } y(x, t) = a \cos(kx - \omega t)$ $\text{Step 1: Compare the given equation with the standard wave equation.}$ $\text{According to the question given wave equation is: } y(x, t) = 0.005 \cos(\alpha x - \beta t)$ $\text{we get: } k = \alpha, \ \text{and } \omega = \beta$ $\text{Step 2: Find the value of the } \alpha \text{ and } \beta.$ $k = \frac{2\pi}{\lambda} \text{ and } \omega = \frac{2\pi}{T}$ $\Rightarrow \frac{2\pi}{\lambda} = \alpha \text{ and } \frac{2\pi}{T} = \beta$ $\text{Given, } \lambda = 0.08 \text{ m and } T = 2.0 \text{ s}$ $\Rightarrow \alpha = \frac{2\pi}{0.08} = 25.00\pi \text{ and } \beta = \frac{2\pi}{2} = \pi.$ $\text{Therefore, the value of } \alpha \text{ and } \beta \text{ are } \alpha = 25\pi \text{ and } \beta = \pi.$ $\text{Hence, option (1) is the correct answer.}$

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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
}