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Current Question (ID: 20469)

Question:

$\text{A small bar magnet placed with its axis at } 30^\circ \text{ with an external field of } 0.06 \text{ T experiences a torque of } 0.018 \text{ Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:}$

Options:

1. $7.2 \times 10^{-2} \text{ J}$
2. $11.7 \times 10^{-3} \text{ J}$
3. $9.2 \times 10^{-3} \text{ J}$
4. $6.4 \times 10^{-2} \text{ J}$

Solution:

$\text{Hint: } W = \Delta U$ $\text{Torque on a bar magnet: } I = MB \sin \theta$ $\text{Here, } \theta = 30^\circ, I = 0.018 \text{ N-m}, B = 0.06 \text{ T}$ $\Rightarrow 0.018 = M \times 0.06 \times \sin 30^\circ$ $\Rightarrow 0.018 = M \times 0.06 \times \frac{1}{2}$ $\Rightarrow M = 0.6 \text{ A-m}^2$ $\text{Now } v = -MB \cos \theta$ $\text{Position of stable equilibrium (} \theta = 0^\circ\text{): } u_1 = -MB$ $\text{Position of unstable equilibrium (} \theta = 180^\circ\text{): } u_i = MB$ $\Rightarrow \text{work done: } \Delta U$ $W = 2MB$ $W = 2 \times 0.6 \times 0.06$ $W = 7.2 \times 10^{-2} \text{ J}$

Import JSON File

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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
}