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Current Question (ID: 17482)

Question:
$\text{A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is:}$
Options:
  • 1. $10 \text{ cm}$
  • 2. $15 \text{ cm}$
  • 3. $2.5 \text{ cm}$
  • 4. $5 \text{ cm}$
Solution:
$\text{By mirror formula, image distance of A}$ $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v_A} + \frac{1}{- (20 + 10)} = \frac{1}{-10}$ $v_A = -15 \text{ cm}$ $\text{Also, image distance of C}$ $v_C = -20 \text{ cm}$ $\text{The length of image} = |v_A - v_C|$ $= |-15 - (-20)|$ $= 5 \text{ cm}$

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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
}