Import Question JSON

Current Question (ID: 18990)

Question:

$\text{The one division of the main scale of vernier callipers reads 1 mm and 10 divisions of the Vernier scale are equal to the 9 divisions on the main scale.}$ $\text{When the two jaws of the instrument touch each other the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division.}$ $\text{When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies between 4.1 cm and 4.2 cm and 6}^{\text{th}} \text{ Vernier division coincides with a main scale division.}$ $\text{The diameter of the bob will be:}$

Options:

1. $1.24 \times 10^{-2} \text{ cm}$
2. $2.40 \times 10^{-2} \text{ cm}$
3. $2.00 \times 10^{-2} \text{ cm}$
4. $4.12 \times 10^{-2} \text{ cm}$

Solution:

$\text{Ans 412}$

Import JSON File

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
}