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Current Question (ID: 8786)

Question:
$\text{In a vernier calliper, } N \text{ divisions of vernier scale coincide with } (N-1) \text{ divisions of the main scale}$ $\text{(in which the length of one division is 1 mm). The least count of the instrument should be:}$
Options:
  • 1. $N \text{ mm}$
  • 2. $(N-1) \text{ mm}$
  • 3. $\frac{1}{10N} \text{ cm}$
  • 4. $\frac{1}{(N-1)} \text{ mm}$
Solution:
$\text{Hint: Least Count = 1 MSD - 1 VSD}$ $\text{Step: Find the least count of the instrument.}$ $N \text{ VSD} = (N-1) \text{ MSD}$ $1 \text{ VSD} = \frac{N-1}{N} \text{ MSD}$ $\text{The least count of the instrument is given by;}$ $\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD}$ $\text{L.C.} = 1 \text{ MSD} - \frac{N-1}{N} \text{ MSD} = \frac{1}{N} \text{ MSD} = \frac{1}{10N} \text{ cm}$ $\text{[As the length of the main scale division is = 1 mm = 0.1 cm]}$ $\text{Therefore, the least count of the instrument is } \frac{1}{10N} \text{ cm.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}