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Current Question (ID: 18930)

Question:
$\text{The distance of the Sun from Earth is } 1.5 \times 10^{11} \text{ m and its angular diameter is } (2000) \text{ s when observed from the earth. The diameter of the Sun will be:}$
Options:
  • 1. $2.45 \times 10^{10} \text{ m}$
  • 2. $1.45 \times 10^{10} \text{ m}$
  • 3. $1.45 \times 10^{9} \text{ m}$
  • 4. $0.14 \times 10^{9} \text{ m}$
Solution:
$\text{Hint: } \theta = \frac{d}{r}$ $\text{Step 1: Convert } (2000) \text{ s in radian.}$ $2000 \text{ s} = \frac{2000}{60 \times 60} \times \frac{\pi}{180}$ $\text{Step 2: Find the diameter of the Sun.}$ $\theta = \frac{d}{r}$ $\Rightarrow d = r \theta = 1.5 \times 10^{11} \times \frac{2000}{60 \times 60} \times \frac{\pi}{180}$ $= 1.45 \times 10^{9} \text{ m}$

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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
}