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Current Question (ID: 16894)

Question:

$\text{If the work done in turning a magnet of magnetic moment } M \text{ by an angle of } 90^\circ \text{ from magnetic meridian is } n \text{ times the corresponding work done to turn it through an angle of } 60^\circ, \text{ then the value of } n \text{ is:}$

Options:

1. $\frac{1}{2}$
2. $2$
3. $\frac{\sqrt{3}}{2}$
4. $\frac{2}{\sqrt{3}}$

Solution:

$\text{Hint: } W = MB(\cos \theta_1 - \cos \theta_2)$ $\text{Step 1: Find the work done in both cases.}$ $W_D = U_f - U_i$ $= PE(\cos \theta_i - \cos \theta_f)$ $W_1 = PE(\cos 0^\circ - \cos 90^\circ) = PE - (1)$ $W_2 = PE(\cos 0^\circ - \cos 60^\circ) = \frac{PE}{2} - (2)$ $\text{Step 2: Find the value of } n.$ $\text{According to the given condition-}$ $W_1 = nW_2$ $PE = n\left(\frac{PE}{2}\right)$ $\Rightarrow n = 2$

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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
}