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Current Question (ID: 20462)

Question:

$\text{A bar magnet with a magnetic moment of } 5 \ \text{Am}^2 \text{ is initially in a stable equilibrium within a uniform external magnetic field of } 0.4 \ \text{T.}$ $\text{The work required to slowly rotate the bar magnet into a position of unstable equilibrium is:}$

Options:

1. $1 \ \text{J}$
2. $2 \ \text{J}$
3. $3 \ \text{J}$
4. $4 \ \text{J}$

Solution:

$\text{Hint: } W = \mu B (\cos \theta_2 - \cos \theta_1)$ $\text{Step: Find the work done in slowly rotating the bar magnet.}$ $\text{The work done in rotating a bar magnet from a stable equilibrium to an unstable equilibrium in a uniform magnetic field is given by:}$ $\Rightarrow W = \mu B (\cos \theta_2 - \cos \theta_1)$ $\text{At stable equilibrium, the magnetic moment is aligned with the magnetic field } (\theta_1 = 0^\circ), \text{ and at unstable equilibrium, the magnetic moment is anti-aligned with the magnetic field } (\theta_2 = 180^\circ).$ $\Rightarrow W = \mu B (\cos \theta_2 - \cos \theta_1)$ $\text{Substitute the given values we get:}$ $\Rightarrow W = 5 \times 0.4 \times (-1 - 1) = 5 \times 0.4 \times (-2) = -4 \ \text{J}$ $\text{The negative sign indicates that work is done against the magnetic force.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}