Import Question JSON

Current Question (ID: 19953)

Question:
$\text{A block of mass } m \text{ attached to a massless spring is performing}$ $\text{oscillatory motion of amplitude } 'A' \text{ on a frictionless horizontal plane.}$ $\text{If half of the mass of the block breaks off when it is passing through}$ $\text{its equilibrium point, the amplitude of oscillation for the remaining}$ $\text{system become } fA. \text{ The value of } f \text{ is:}$
Options:
  • 1. $\sqrt{2}$
  • 2. $1$
  • 3. $\frac{1}{2}$
  • 4. $\frac{1}{\sqrt{2}}$
Solution:
$\text{Hint: Velocity is maximum at the equilibrium position.}$ $\text{At equilibrium position}$ $V_0 = \omega_1 A = \sqrt{\frac{K}{m}} A$ $V = \omega A^1 = \sqrt{\frac{K}{m/2}} A^1$ $\therefore A^1 = \frac{A}{\sqrt{2}}$

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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
}