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Current Question (ID: 16430)

Question:
$\text{In the following four situations, charged particles are at an equal distance from the origin.}$ $\text{Arrange the magnitude of the net electric field at origin, starting with the highest.}$
Options:
  • 1. $(i) > (ii) > (iii) > (iv)$
  • 2. $(ii) > (i) > (iii) > (iv)$
  • 3. $(i) > (iii) > (ii) > (iv)$
  • 4. $(iv) > (iii) > (ii) > (i)$
Solution:
$\text{Hint: Use } R = \sqrt{A^2 + B^2 + 2AB \cos \theta}. $ $\text{Step: Find the magnitude of the electric field in descending order.}$ $\text{If the electric field due to charge } |q| \text{ at origin is } E \text{ then the electric field due to charges } |2q|, |3q|, |4q| \text{ and } |5q| \text{ are respectively } 2E, 3E, 4E \text{ and } 5E.$ $\text{E}_{(i)} = \sqrt{(5E)^2 + (5E)^2} = 5\sqrt{2}E,$ $\text{E}_{(ii)} = \sqrt{(3E)^2 + (3E)^2} = 3\sqrt{2}E,$ $\text{E}_{(iii)} = 4E + 2E = 6E \text{ and } \text{E}_{(iv)} = 3E + E = 4E$ $\Rightarrow \text{E}_{(i)} > \text{E}_{(iii)} > \text{E}_{(ii)} > \text{E}_{(iv)}$ $\text{Hence, option (3) is the correct answer.}$

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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
}