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Current Question (ID: 11268)

Question:

$\text{Two simple pendulums of length 1 m and 16 m are in the same phase at the mean position at any instant. If } T \text{ is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:}$

Options:

1. $\frac{3T}{2}$
2. $\frac{3T}{4}$
3. $\frac{2T}{3}$
4. $\frac{4T}{3}$ (Correct)

Solution:

$\text{The time period of the smaller pendulum } = T = 2\pi \sqrt{\frac{l}{g}}$ $\text{The time period of the larger pendulum } = 2\pi \sqrt{\frac{16}{g}} = 4 \times 2\pi \sqrt{\frac{l}{g}} = 4T$ $\text{At any time } t, \phi_1 = \frac{2\pi}{T} t \text{ & } \phi_2 = \frac{2\pi}{4T} t$ $\text{They will be in phase again when the faster pendulum completes one oscillation with respect to the slower one.}$ $\Rightarrow \phi_1 - \phi_2 = 2\pi$ $\Rightarrow \frac{2\pi}{T} t - \frac{2\pi}{4T} \cdot t = 2\pi \Rightarrow t = \frac{4T}{3}$ $\text{Note: The pendulums have the same phase when they are at the same point and are moving in the same direction.}$

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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
}