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Current Question (ID: 16011)

Question:

$\text{Three waves of equal frequency having amplitudes of } 10 \, \mu m, 4 \, \mu m \text{ and } 7 \, \mu m \text{ arrive at a given point with a successive phase difference of } \frac{\pi}{2}. \text{ The amplitude of the resulting wave (in } \mu m\text{) is given by:}$

Options:

1. $7$
2. $6$
3. $5$
4. $4$

Solution:

$\text{Hint: } R = \sqrt{A^2 + B^2}$ $\text{Step: Find the amplitude of the resultant wave.}$ $\text{The wave 1 and 3 reach out of phase. Hence the resultant phase difference between them is } \pi.$ $\text{The resultant amplitude of (1) and (3) are } R = 10 - 7 = 3 \, \mu m$ $\text{This wave has a phase difference of } \frac{\pi}{2} \text{ with } 4 \, \mu m$ $\text{The resultant amplitude } R = \sqrt{3^2 + 4^2} = 5 \, \mu m$ $\text{Hence, option (3) is the correct answer.}$

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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
}