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Current Question (ID: 17001)

Question:

$\text{A conducting rod } AC \text{ of length } 4l \text{ is rotated about point } O \text{ in a uniform magnetic field } \vec{B} \text{ directed into the paper.}$ $\text{If } AO = l \text{ and } OC = 3l, \text{ then:}$

Options:

1. $V_A - V_O = \frac{B \omega l^2}{2}$
2. $V_O - V_C = \frac{7}{2} B \omega l^2$
3. $V_A - V_C = 4 B \omega l^2$
4. $V_C - V_O = \frac{-1}{2} B \omega l^2$

Solution:

$\text{(3) By using } V = \frac{1}{2} B l^2 \omega$ $\text{For part } AO; \ V_{OA} = V_O - V_A = \frac{1}{2} B l^2 \omega$ $\text{For part } OC; \ V_{OC} = V_O - V_C = \frac{1}{2} B (3l)^2 \omega$ $\therefore \ V_A - V_C = 4 B l^2 \omega$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}