The main scale of a vernier calliper has $n$ divisions/cm. $n$ divisions of the vernier scale coincide with $(

Question

The main scale of a vernier calliper has $n$ divisions/cm. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier calliper is:

Accepted Answer

1. **Find the value of 1 Main Scale Division (MSD):** Since there are $n$ divisions per cm on the main scale, the length of $1 	ext{ MSD} = \frac{1}{n}	ext{ cm}$.
2. **Find the relation between MSD and VSD:** It is given that $n$ divisions of the vernier scale (VSD) coincide with $(n-1)$ divisions of the main scale (MSD). Therefore, $n 	ext{ VSD} = (n-1) 	ext{ MSD}$, which implies $1 	ext{ VSD} = \frac{n-1}{n} 	ext{ MSD}$.
3. **Calculate the Least Count (LC):** The least count of a vernier calliper is defined as the difference between one main scale division and one vernier scale division.
$	ext{LC} = 1 	ext{ MSD} - 1 	ext{ VSD}$
$	ext{LC} = 1 	ext{ MSD} - \frac{n-1}{n} 	ext{ MSD} = \left(1 - \frac{n-1}{n}ight) 	ext{ MSD} = \frac{1}{n} 	ext{ MSD}$.
4. **Substitute the value of 1 MSD:**
$	ext{LC} = \frac{1}{n} 	imes \left(\frac{1}{n}	ext{ cm}ight) = \frac{1}{n^2}	ext{ cm}$.

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