Given:
- 20 vernier scale divisions (VSD) coincide with 19 main scale divisions (MSD).
- Least count (L.C.) of the instrument is \( 0.1 \, \text{mm} \).
Step 1: Relation Between VSD and MSD
From the given data:
\[ 20 \, \text{VSD} = 19 \, \text{MSD}. \]
The value of one vernier scale division (1 VSD) is:
\[ 1 \, \text{VSD} = \frac{19}{20} \, \text{MSD}. \]
Step 2: Calculating the Least Count
The least count (L.C.) is given by:
\[ \text{L.C.} = 1 \, \text{MSD} - 1 \, \text{VSD}. \]
Substituting the value of 1 VSD:
\[ \text{L.C.} = 1 \, \text{MSD} - \frac{19}{20} \, \text{MSD} = \frac{1}{20} \, \text{MSD}. \]
Given that the least count is \( 0.1 \, \text{mm} \):
\[ 0.1 \, \text{mm} = \frac{1}{20} \, \text{MSD}. \]
Multiplying both sides by 20:
\[ 1 \, \text{MSD} = 2 \, \text{mm}. \]
Therefore, one main scale division is equal to \( 2 \, \text{mm} \).
If \( T = 2\pi \sqrt{\frac{L}{g}} \), \( g \) is a constant and the relative error in \( T \) is \( k \) times to the percentage error in \( L \), then \( \frac{1}{k} = \) ?