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Current Question (ID: 19913)

Question:

$\text{The time period of a simple pendulum in a stationary lift is } T. \text{ If the lift accelerates upward with an acceleration of } \frac{g}{6} \text{ (where } g \text{ is the acceleration due to gravity), then the time period of the pendulum would be:}$

Options:

1. $\sqrt{\frac{6}{5}} T$
2. $\sqrt{\frac{5}{6}} T$
3. $\sqrt{\frac{6}{7}} T$
4. $\sqrt{\frac{7}{6}} T$

Solution:

$\text{Hint: } T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}}$ $\text{Step: Find the time period of the simple pendulum.}$ $\text{The time period of the simple pendulum in a stationary lift is given by:}$ $T = 2\pi \sqrt{\frac{l}{g}} \quad \ldots (1)$ $\text{The time period of the simple pendulum in a lift accelerates upwards is given by:}$ $\Rightarrow T' = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} \quad \ldots (2)$ $\text{The effective acceleration of the pendulum is given by:}$ $g_{\text{eff}} = g + \frac{g}{6} = \frac{7}{6} g$ $\text{Divide equation (2) by (1) we get:}$ $\Rightarrow \frac{T'}{T} = \sqrt{\frac{6}{7}}$ $\Rightarrow T' = \sqrt{\frac{6}{7}} T$ $\text{Hence, option (3) is the correct answer.}$

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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
}