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Current Question (ID: 17515)

Question:
$\text{In the following diagram, what is the distance } x \text{ if the radius of curvature is } R = 15 \text{ cm?}$
Options:
  • 1. $30 \text{ cm}$
  • 2. $20 \text{ cm}$
  • 3. $15 \text{ cm}$
  • 4. $10 \text{ cm}$
Solution:
$\text{Hint: } \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}$ $\text{Step: Find the distance } x \text{ in the following diagram.}$ $\text{Take } u, v \text{ and } R \text{ according to the sign convention}$ $u = -x , v = \infty , R = 15$ $\text{Use the formula for refraction at the curved interface}$ $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2-\mu_1}{R}$ $\frac{1.5}{\infty} + \frac{1}{x} = \frac{1.5-1}{15}$ $\Rightarrow x = 30 \text{ cm}$ $\text{Hence, option (1) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}