Import Question JSON

Current Question (ID: 18925)

Question:

$\text{The pitch of the screw gauge is } 1 \text{ mm and there are } 100 \text{ divisions on the circular scale.}$ $\text{When nothing is put in between the jaws, the zero of the circular scale lies } 8 \text{ divisions below the reference line.}$ $\text{When a wire is placed between the jaws, the first linear scale division is clearly visible while the } 72^{\text{nd}} \text{ division on a circular scale coincides with the reference line.}$ $\text{The radius of the wire is:}$

Options:

1. $1.64 \text{ mm}$
2. $0.82 \text{ mm}$
3. $1.80 \text{ mm}$
4. $0.90 \text{ mm}$

Solution:

$\text{Hint: Least count = pitch / no. of circular divisions}$ $\text{Step 1: Find the least count of screw gauge.}$ $\text{L.C. = pitch / No. of C.S.D}$ $= 1 \text{ mm} / 100$ $= 0.01 \text{ mm}$ $\text{Step 2: Find the zero error.}$ $\text{Positive error} = \text{L.C.} \times \text{divisions}$ $= 0.01 \times 8$ $= 0.08 \text{ mm}$ $\text{Step 3: Find the observed reading.}$ $\text{O.R.} = \text{M.S.R.} + \text{C.S.R.}$ $= 1 \text{ mm} + 72 \times 0.01 \text{ mm}$ $= 1.72 \text{ mm}$ $\text{Step 4: Find the current reading.}$ $\text{C.R.} = \text{O.B.} - \text{zero error}$ $= 1.72 - (0.08) = 1.64 \text{ mm}$ $\text{Step 5: Find the radius}$ $r = d/2 = 1.64/2 = 0.82 \text{ mm.}$

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
}