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Current Question (ID: 17535)

Question:

$\text{In the diagram shown below, the image of the point object } O \text{ is formed at } I \text{ by the convex lens of focal length } 20 \text{ cm, where } F_1 \text{ and } F_2 \text{ are foci of the lens.}$ $\text{The value of } x' \text{ is:}$

Options:

1. $10 \text{ cm}$
2. $20 \text{ cm}$
3. $30 \text{ cm}$
4. $40 \text{ cm}$

Solution:

$\text{Hint: } \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\text{Step: Find } 'u' \text{ and } 'v' \text{ according to the sign convention.}$ $u = -(30) \text{ cm}$ $v = (20 + x') \text{ cm}$ $\text{Using the lens formula we get;}$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{20 + x'} + \frac{1}{30} = \frac{1}{20}$ $x' = 40 \text{ cm}$ $\text{Hence, option (4) is the correct answer.}$

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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
}