Given:
\[ 10 \, \text{MSD} = 11 \, \text{VSD} \]
1 VSD (Vernier Scale Division) is equivalent to:
\[ 1 \, \text{VSD} = \frac{10}{11} \, \text{MSD} \]
The least count (LC) of the Vernier caliper is given by:
\[ LC = 1 \, \text{MSD} - 1 \, \text{VSD} \]
Substituting the values:
\[ LC = 1 \, \text{MSD} - \frac{10}{11} \, \text{MSD} = \frac{1}{11} \, \text{MSD} \]
Given that 1 MSD corresponds to 5 units:
\[ LC = \frac{5 \, \text{units}}{11} \]
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to: