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Current Question (ID: 20076)

Question:
$\text{Two point charges } q_1(\sqrt{10} \ \mu\text{C}) \text{ and } q_2(-25 \ \mu\text{C}) \text{ are placed on the } x\text{-axis at } x = 1 \ \text{m and } x = 4 \ \text{m respectively.}$ $\text{The electric field (in V/m) at a point } y = 3 \ \text{m on the } y\text{-axis is:}$ $\left(\text{take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \ \text{Nm}^2\text{C}^{-2}\right)$
Options:
  • 1. $(63\hat{i} - 27\hat{j}) \times 10^2$
  • 2. $(-63\hat{i} + 27\hat{j}) \times 10^2$
  • 3. $(81\hat{i} - 81\hat{j}) \times 10^2$
  • 4. $(-81\hat{i} + 81\hat{j}) \times 10^2$
Solution:
$\text{Hint: } E = \frac{kq}{r^2}$

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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
}