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Current Question (ID: 19939)

Question:

$\text{In the context of linear simple harmonic motion (SHM), consider the following statements:}$ $\text{(A) Acceleration is maximum at the mean position.}$ $\text{(B) Velocity is maximum at the extreme position.}$ $\text{(C) Acceleration is maximum at the extreme position.}$ $\text{(D) Velocity is maximum at the mean position.}$ $\text{Choose the correct option from the given ones:}$

Options:

1. $(\text{B}), (\text{C}) \text{ and } (\text{D}) \text{ only}$
2. $(\text{A}) \text{ and } (\text{D}) \text{ only}$
3. $(\text{A}) \text{ and } (\text{B}) \text{ only}$
4. $(\text{C}) \text{ and } (\text{D}) \text{ only}$

Solution:

$\text{Hint: } |a| = \omega^2 x \text{ and } v = \omega \sqrt{A^2 - x^2}$ $\text{Step: Analyse each statement one by one.}$ $\text{(A) The acceleration of the particle performing simple harmonic motion (SHM) is given by:}$ $a = -\omega^2 x$ $\text{At the mean position i.e., } x = 0, \text{ the acceleration of the particle is zero.}$ $\text{Hence, statement (A) is False.}$ $\text{(B) The velocity of the particle performing simple harmonic motion (SHM) is given by:}$ $v = \omega \sqrt{A^2 - x^2}$ $\text{At the extreme position i.e., } x = \pm A, \text{ the velocity of the particle is zero.}$ $\text{Hence, statement (B) is False.}$ $\text{(C) At the extreme position i.e., } x = \pm A, \text{ the acceleration of the particle is maximum } (a = -\omega^2 A).$ $\text{Hence, statement (C) is True.}$ $\text{(D) At the mean position i.e., } x = 0, \text{ the velocity of the particle is maximum } (v = A \omega).$ $\text{Hence, statement (D) is True.}$ $\text{Therefore, statements (C) and (D) are True.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}