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Current Question (ID: 16446)

Question:
$\text{An electric dipole is kept at the origin as shown in the diagram. The point } A, B, C \text{ are on a circular arc with the centre of curvature at the origin. If the electric fields at } A, B \text{ and } C \text{ respectively are } \vec{E}_1, \vec{E}_2, \vec{E}_3, \text{ then which of the following is incorrect? } (d \gg l)$
Options:
  • 1. $\vec{E}_1 = -\vec{E}_3$
  • 2. $\vec{E}_1 = -2\vec{E}_2$
  • 3. $\vec{E}_1 = \vec{E}_3$
  • 4. $\vec{E}_3 = -2\vec{E}_2$
Solution:
$\vec{E}_{\text{axial}} = \frac{2k\vec{p}}{r^3}, \vec{E}_{\text{equatorial}} = \frac{-k\vec{p}}{r^3}$ $\text{Step 1: Identify point dipole}$ $\text{Step 2: Find the electric field for axial and equatorial position}$ $\vec{E}_{\text{ax}} = \frac{2k\vec{p}}{r^3}, \vec{E}_{\text{eq}} = \frac{-k\vec{p}}{r^3}$ $\vec{E}_1 = \vec{E}_3 = \frac{2k\vec{p}}{d^3}$ $\vec{E}_2 = \frac{-k\vec{p}}{d^3}$

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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
}