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Current Question (ID: 12060)

Question:

In a vernier callipers, one main scale division is x cm and n division of the vernier scale coincide with (n-1) divisions of the main scale. The least count (in cm) of the callipers is

Options:

1. $(\frac{n-1}{n})x$
2. $\frac{nx}{(n-1)}$
3. $\frac{x}{n}$
4. $\frac{x}{(n-1)}$

Solution:

Least Count (LC) of a vernier caliper is given by the formula: $LC = 1\text{ MSD} - 1\text{ VSD}$. Given: $1\text{ MSD} = x\text{ cm}$. We are also given that $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. Therefore, $n\text{ VSD} = (n-1)\text{ MSD}$. $1\text{ VSD} = \frac{(n-1)}{n}\text{ MSD} = \frac{(n-1)}{n}x\text{ cm}$. Substituting this into the LC formula: $LC = x - \frac{(n-1)}{n}x = x(1 - \frac{n-1}{n}) = x(\frac{n-(n-1)}{n}) = x(\frac{n-n+1}{n}) = \frac{x}{n}\text{ cm}$.

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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
}