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Current Question (ID: 20029)

Question:
$\text{Charges } Q_1 \text{ and } Q_2 \text{ are at points } A \text{ and } B \text{ of a right angle triangle } OAB \text{ (see figure).}$ $\text{The resultant electric field at point } O \text{ is perpendicular to the hypotenuse, then } \frac{Q_1}{Q_2} \text{ is proportional to:}$
Options:
  • 1. $\frac{x_2}{x_1^2}$
  • 2. $\frac{x_2^2}{x_1}$
  • 3. $\frac{x_2^3}{x_1^3}$
  • 4. $\frac{x_1^3}{x_2^3}$
Solution:
$\text{Hint: } E = \frac{kQ}{R^2}$ $E_2 = \text{electric field due to } Q_2 = \frac{kQ_2}{x_2^2}$ $E_1 = \frac{kQ_1}{x_1^2}$ $\text{From diagram}$ $\tan \theta = \frac{E_2}{E_1} = \frac{x_1}{x_2}$ $\frac{kQ_2}{x_2^2} \times \frac{x_1^2}{kQ_1} = \frac{x_1}{x_2}$ $Q_2 x_1^2 = Q_1 x_2^2$ $\frac{Q_2}{Q_1} = \frac{x_2^2}{x_1^2}$ $\frac{Q_1}{Q_2} = \frac{x_1^2}{x_2^2}$

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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
}