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Current Question (ID: 15982)

Question:

$\text{Two waves represented by the following equations are travelling in the same medium } y_1 = 5 \sin 2\pi (75t - 0.25x), y_2 = 10 \sin 2\pi (150t - 0.50x).$ $\text{The intensity ratio } \frac{I_1}{I_2} \text{ of the two waves will be:}$

Options:

1. $1 : 2$
2. $1 : 4$
3. $1 : 8$
4. $1 : 16$

Solution:

$\text{Hint: } I \propto \omega^2 A^2$ $\text{Step: Find the intensity ratio } \frac{I_1}{I_2} \text{ of two waves.}$ $\text{The intensity of the resultant wave is proportional to } \omega^2 A^2$ $\text{The two waves are represented by: } y_1 = 5 \sin 2\pi (75t - 0.25x) \text{ and } y_2 = 10 \sin 2\pi (150t - 0.50x).$ $\text{Given: } \omega_1 = 2\pi \times 75 = 150\pi, \omega_2 = 2\pi \times 150 = 300\pi \text{ and amplitudes are } A_1 = 5, A_2 = 10$ $\text{So, the ratio of the intensity } \frac{I_1}{I_2} \text{ is}$ $\frac{I_1}{I_2} \propto \left( \frac{\omega_1}{\omega_2} \right)^2 \left( \frac{A_1}{A_2} \right)^2 \propto \left( \frac{150\pi}{300\pi} \right)^2 \times \left( \frac{5}{10} \right)^2 \propto \left( \frac{1}{2} \right)^4 \propto \frac{1}{16}$ $\text{Therefore, the intensity ratio } \frac{I_1}{I_2} \text{ of the two waves is } \frac{1}{16}.$ $\text{Hence, option (4) is the correct answer.}$

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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
}