Import Question JSON

Current Question (ID: 18938)

Question:

$\text{In an experiment to determine the diameter of a wire using a screw gauge, the following observations were recorded:}$ $\text{(a) The screw moves 0.5 mm on the main scale in one complete rotation.}$ $\text{(b) Total divisions on circular scale = 50.}$ $\text{(c) The main scale reading is 2.5 mm.}$ $\text{(d) 45}^{\text{th}} \text{ division of the circular scale is in the pitch line.}$ $\text{(e) The instrument has 0.03 mm negative error.}$ $\text{Then the diameter of the wire is:}$

Options:

1. $2.92 \text{ mm}$
2. $2.54 \text{ mm}$
3. $2.98 \text{ mm}$
4. $3.45 \text{ mm}$

Solution:

$\text{Hint: Actual diameter = measured diameter - zero error.}$ $\text{Step 1: Find the circular scale reading.}$ $\text{C.S.R = 45} \times \text{Least count}$ $\Rightarrow \text{C.S.R = 45} \times \frac{\text{Pitch}}{\text{Divisions on circular scale}}$ $\Rightarrow \text{C.S.R = 45} \times \frac{0.5}{50}$ $\Rightarrow \text{C.S.R = 0.45 mm}$ $\text{Step 2: Find the diameter of the wire.}$ $\text{T.R. = M.S.R + C.S.R + Negative error}$ $\Rightarrow \text{T.R. = 2.5 + 0.45 + 0.03}$ $\Rightarrow \text{T.R. = 2.98 mm}$ $\text{Hence, option (3) is the correct answer.}$

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
}