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Current Question (ID: 17475)

Question:
$\text{An object is at a distance of } 30 \text{ cm in front of a concave mirror of focal length } 10 \text{ cm. The image of the object will be:}$
Options:
  • 1. $\text{smaller in size.}$
  • 2. $\text{inverted.}$
  • 3. $\text{between the focus and centre of curvature.}$
  • 4. $\text{All of the above.}$
Solution:
$\text{Hint: magnification, } m = \frac{f}{f-u}$ $\text{Step 1: Write object distance and focal length according to sign convention.}$ $u = -30 \text{ cm}$ $f = -10 \text{ cm}$ $\text{Step 2: Use the magnification formula.}$ $m = \frac{f}{f-u} = \frac{-10}{-10 - (-30)}$ $\Rightarrow -\frac{v}{u} = \frac{-1}{2}$ $\Rightarrow v = \frac{u}{2}$ $= -15 \text{ cm}$ $\text{Step 3: Find the nature of the image.}$ $\text{Here, } m = \text{(negative)}$ $\text{Thus, the image will be inverted, and small and between the focus and centre of curvature.}$

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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
}