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Current Question (ID: 19904)

Question:
$\text{In the given figure, mass } M \text{ is attached to a horizontal spring which is fixed on one side to a rigid support.}$ $\text{The spring constant of the spring is } k. \text{ The mass oscillates on a frictionless surface with time period } T \text{ and amplitude } A.$ $\text{When the mass is in an equilibrium position, as shown in the figure, another mass } m \text{ is gently fixed upon it.}$ $\text{The new amplitude of oscillation will be:}$
Options:
  • 1. $A \sqrt{\frac{M-m}{M}}$
  • 2. $A \sqrt{\frac{M}{M+m}}$
  • 3. $A \sqrt{\frac{M+m}{M}}$
  • 4. $A \sqrt{\frac{M}{M-m}}$
Solution:
$\text{Hint: Momentum of system remains conserved.}$ $p_i = p_f$ $MA \omega = (m+M) A' \omega'$ $MA \sqrt{\frac{k}{M}} = (m+M) A' \sqrt{\frac{k}{m+M}}$ $A' = A \sqrt{\frac{M}{M+m}}$

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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
}