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Current Question (ID: 18980)

Question:

$\text{A travelling microscope is used to determine the refractive index of a glass slab. If } 40 \text{ divisions are there in } 1 \text{ cm on the main scale and } 50 \text{ Vernier scale divisions are equal to } 49 \text{ main scale divisions, then the least count of the travelling microscope is:}$

Options:

1. $3 \times 10^{-6} \text{ m}$
2. $4 \times 10^{-6} \text{ m}$
3. $5 \times 10^{-6} \text{ m}$
4. $6 \times 10^{-6} \text{ m}$

Solution:

$\text{Least count} = 1 \text{ MSD} - 1 \text{ VSD}$ $\text{MSD} = \frac{1}{40} \text{ cm}$ $50 \text{ VSD} = 49 \text{ MSD}$ $1 \text{ VSD} = \frac{49}{50} \times \frac{1}{40} \text{ cm}$ $\text{Least count} = \frac{1}{40} - \frac{49}{50 \times 40} = \frac{1}{200} \text{ cm} = 5 \times 10^{-6} \text{ m}$

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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
}