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Current Question (ID: 19085)

Question:

$\text{A clock has a continuously moving second's hand of } 0.1 \text{ m length.}$ $\text{The average acceleration of the tip of the second hand (in units of } \text{ms}^{-2}) \text{ is of the order of:}$

Options:

1. $10^{-3}$
2. $10^{-4}$
3. $10^{-1}$
4. $10^{-2}$

Solution:

$\text{Hint: } a = \omega^2 R$ $\text{Step: Find the order of the average acceleration of the particle.}$ $\text{The length of the minute hand is } 0.1 \text{ m.}$ $\text{The angular frequency of the minute hand is given by:}$ $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = 0.105 \text{ rad/s}$ $\text{The average acceleration of the tip is given by:}$ $a = \omega^2 R = (0.105)^2 (0.1) = 0.0011 = 1.1 \times 10^{-3}$ $\text{Therefore, the average acceleration is of the order of } 10^{-3}.$ $\text{Hence, option (1) is the correct answer.}$

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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
}