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Current Question (ID: 9877)

Question:

$\text{The potential energy of a particle in a force field is } U = \frac{A}{r^2} - \frac{B}{r} \text{ where A and B are positive constants and r is the distance of the particle from the centre of the field. For stable equilibrium, the distance of the particle is:}$

Options:

1. $\frac{B}{A}$
2. $\frac{B}{2A}$
3. $\frac{2A}{B}$
4. $\frac{A}{B}$

Solution:

\text{Hint: } F = -\frac{dU}{dr} \text{Step: Find the distance of the particle from centre of the field.} \text{Given: The potential energy of the particle in a force field is:} U = \frac{A}{r^2} - \frac{B}{r} \text{For stable equilibrium, } F = -\frac{dU}{dr} = 0 \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} = 0 B = \frac{2A}{r} \text{The distance of the particle from centre of the field is given by:} r = \frac{2A}{B} \text{Hence, option (3) 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
}