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Current Question (ID: 11298)

Question:
$\text{The displacement } x \text{ of a particle varies with time } t \text{ as } x = A\sin\left(\frac{2\pi t}{T} + \frac{\pi}{3}\right)$ $\text{. The time taken by the particle to reach from } x = \frac{A}{2} \text{ to } x = -\frac{A}{2} \text{ will be:}$
Options:
  • 1. $\frac{T}{2}$
  • 2. $\frac{T}{3}$
  • 3. $\frac{T}{12}$
  • 4. $\frac{T}{6}$ (Correct)
Solution:
$\text{Hint: Use a phasor diagram to find the time taken by a particle to reach from } x = \frac{A}{2} \text{ to } x = -\frac{A}{2}.$ $\text{Step 1: Draw the phasor diagram.}$ $\text{Step 2: Find the angle with the help of the diagram.}$ $A\sin\theta = \frac{A/2}{A} = \frac{1}{2}$ $\Rightarrow \theta = 30°$ $\Rightarrow \Delta\theta = \frac{\pi}{3}.$ $\text{Step 3: Find the time taken by a particle to reach from } x = \frac{A}{2} \text{ to } x = -\frac{A}{2}.$ $t = \frac{\Delta\theta}{\omega} \text{ where, } \omega = \frac{2\pi}{T}$ $\Rightarrow t = \frac{T}{6}.$ $\text{Option (4) is correct.}$

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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
}