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Current Question (ID: 19005)

Question:

$\text{While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is } 1 \text{ mm, and the circular scale reading is equal to } 42 \text{ divisions. The pitch of the screw gauge is } 1 \text{ mm, and it has } 100 \text{ divisions on the circular scale. The diameter of the wire is } \frac{x}{50} \text{ mm. The value of } x \text{ is:}$

Options:

1. $142$
2. $21$
3. $42$
4. $72$

Solution:

$\text{The total reading of the screw gauge is given by:}$ $\text{Total reading} = \text{Main scale reading} + \left(\frac{\text{Circular scale reading}}{\text{Total divisions}} \times \text{Pitch}\right)$ $= 1 + \left(\frac{42}{100} \times 1\right) = 1.42 \text{ mm}$ $\text{Given that the diameter of the wire is } \frac{x}{50} \text{ mm, we equate:}$ $\frac{x}{50} = 1.42$ $x = 1.42 \times 50 = 71$

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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
}