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Current Question (ID: 9876)

Question:
$\text{A particle is moving such that the potential energy U varies with position in metre as } U(x) = (4x^2 - 2x + 50) \text{ J. The particle will be in equilibrium at:}$
Options:
  • 1. $x = 25 \text{ cm}$
  • 2. $x = 2.5 \text{ cm}$
  • 3. $x = 25 \text{ m}$
  • 4. $x = 2.5 \text{ m}$
Solution:
$\text{To find the equilibrium position, set the derivative of the potential energy } U(x) \text{ with respect to } x \text{ to zero.}$ $\text{Given } U(x) = 4x^2 - 2x + 50$ $\frac{dU}{dx} = 8x - 2$ $\text{Set } \frac{dU}{dx} = 0:$ $8x - 2 = 0$ $8x = 2$ $x = \frac{2}{8} = \frac{1}{4} = 0.25 \text{ meters}$ $\text{Convert to centimeters:}$ $x = 0.25 \text{ m} \times 100 \text{ cm/m} = 25 \text{ cm}$ $\text{Thus, the particle is in equilibrium at } x = 25 \text{ cm.}$

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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
}