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Current Question (ID: 15939)

Question:

$\text{The wave described by } y = 0.25 \sin(10\pi x - 2\pi t), \text{ where } x \text{ and } y \text{ are in metres and } t \text{ in seconds, is a wave traveling along the:}$

Options:

1. $-\ve x \text{ direction with frequency 1 Hz.}$
2. $+\ve x \text{ direction with frequency } \pi \text{ Hz and wavelength } \lambda = 0.2 \text{ m.}$
3. $+\ve x \text{ direction with frequency 1 Hz and wavelength } \lambda = 0.2 \text{ m.}$
4. $-\ve x \text{ direction with amplitude 0.25 m and wavelength } \lambda = 0.2 \text{ m.}$

Solution:

$\text{Hint: } k = \frac{2\pi}{\lambda}$ $\text{Step: Analyse each option one by one.}$ $\text{The sign between two terms in the sine argument will define the direction of the wave propagation.}$ $\text{Writing the given wave equation}$ $y = 0.25 \sin(10\pi x - 2\pi t) \quad \cdots (i)$ $\text{The minus (-) between } (10\pi x) \text{ and } (2\pi t) \text{ implies that the wave is travelling along positive x direction.}$ $\text{Now comparing Eq. (i) with the standard wave equation}$ $y = a \sin(kx - \omega t) \quad \cdots (ii)$ $\text{we have}$ $a = 0.25 \text{ m}, \omega = 2\pi, k = 10\pi \text{ m}$ $\therefore \frac{2\pi}{T} = 2\pi$ $\Rightarrow f = 1 \text{ Hz}$ $\text{Also, } \lambda = \frac{2\pi}{k} = \frac{2\pi}{10\pi} = 0.2 \text{ m}$ $\text{Therefore, the wave is travelling along } +\ve x \text{ direction with frequency 1 Hz and wavelength 0.2 m.}$ $\text{Hence, option (3) is the correct answer.}$

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
}