Import Question JSON

Current Question (ID: 20093)

Question:
$\text{The electric field due to a dipole at a point on its equatorial plane varies as } r^{-n}. \text{ The value of } n \text{ is:}$
Options:
  • 1. $0$
  • 2. $2$
  • 3. $-3$
  • 4. $3$
Solution:
$\text{Hint: } \vec{E} = \frac{k p}{r^3}$ $\text{Step: Find the value of } n.$ $\text{At an equatorial point of a dipole, the electric field due to the dipole is given by:}$ $E_{\text{equatorial}} = \frac{k P}{r^3}$ $\Rightarrow E \propto \frac{1}{r^3}$ $\text{Therefore, the dependence of the electric field on the distance } r \text{ at an equatorial point is } \frac{1}{r^3}.$ $\Rightarrow E_{\text{equatorial}} \propto r^{-3} \propto r^{-n}$ $\Rightarrow n = 3$ $\text{Hence, option (4) is the correct answer.}$

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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
}