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Current Question (ID: 18935)

Question:

$\text{In a Vernier Calipers, 10 divisions of the Vernier scale is equal to the 9 divisions of the main scale.}$ $\text{When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4}^{\text{th}} \text{ Vernier scale division exactly coincides with the main scale reading.}$ $\text{One main scale division is equal to 1 mm.}$ $\text{While measuring diameter of a spherical body, the body is held between two jaws.}$ $\text{It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6}^{\text{th}} \text{ Vernier scale division exactly coincides with the main scale reading.}$ $\text{The diameter of the spherical body will be:}$

Options:

1. $3.02 \text{ cm}$
2. $3.06 \text{ cm}$
3. $3.10 \text{ cm}$
4. $3.20 \text{ cm}$

Solution:

$\text{Least Count (LC) of Vernier Calipers} = \text{1 MSD} - \text{1 VSD}$ $= \text{1 MSD} - \frac{9}{10} \text{MSD}$ $= \frac{1}{10} \times \text{MSD} = 0.1 \text{ mm}$ $\text{Diameter} = 30 \times 1 + 6 \times 0.1 = 30 + 0.6 = 30.6 \text{ mm} = 3.06 \text{ cm}$

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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
}