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Current Question (ID: 19912)

Question:

$\text{The time period of a simple pendulum is } T. \text{ The time taken to complete } \frac{5}{8} \text{ oscillations starting from the mean position is } \frac{\alpha}{\beta} T. \text{ The value of } \alpha \text{ is:}$

Options:

1. $3$
2. $6$
3. $7$
4. $10$

Solution:

$\left( \frac{5}{8} \right)^{\text{th}} \text{ of oscillation} = \left( \frac{1}{2} + \frac{1}{8} \right)^{\text{th}} \text{ of oscillation.}$ $\text{Find the value of } \alpha. \text{ The total distance covered by the particle in one time period is } 4A.$ $\frac{5}{8} \text{ of oscillation means } \frac{5}{8} \text{ of total distance covered in 1 complete oscillations i.e., } \frac{5}{8} \times 4A = \frac{5}{2}A.$ $\text{The total time taken by the particle to cover } 2.5A \text{ distance is given by:}$ $t = \frac{T}{4} + \frac{T}{4} + \frac{T}{12} = \frac{7T}{12} \quad \ldots (1)$ $\text{Now, comparing equation (1) with the time given in the question, we get:}$ $\frac{\alpha T}{\beta} = \frac{7T}{12}$ $\Rightarrow \alpha = 7, \beta = 12$ $\text{Therefore, the value of } \alpha \text{ is } 7.$ $\text{Hence, option (3) is the correct answer.}$

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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
}