Import Question JSON

Current Question (ID: 19014)

Question:

$\text{One main scale division of a vernier caliper is equal to } m \text{ units.}$ $\text{If } n\text{th division of main scale coincides with } (n+1)\text{th division of vernier scale,}$ $\text{the least count of the vernier caliper is:}$

Options:

1. $\frac{1}{(n+1)}$
2. $\frac{n}{(n+1)}$
3. $\frac{m}{n(n+1)}$
4. $\frac{m}{(n+1)}$

Solution:

$\text{Least count of vernier caliper} = \text{Main scale division} - \text{Vernier scale division}$ $= m - \frac{m}{n+1}$ $= \frac{m(n+1) - m}{n+1}$ $= \frac{m}{n+1}$

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
}