Vernier Calipers, Micrometer screw gauge Questions in English

(C) The zero of the circular scale lies $3$ divisions above the pitch line,which indicates a negative zero error.
Zero error $e = -3 \times LC = -3 \times 0.01 \ mm = -0.03 \ mm$.
The observed reading is given by: $\text{Pitch scale reading} + (\text{Circular scale reading} \times LC)$.
Observed reading $= 1 \ mm + 51 \times 0.01 \ mm = 1 \ mm + 0.51 \ mm = 1.51 \ mm$.
The correct reading is calculated as: $\text{Correct reading} = \text{Observed reading} - \text{Zero error}$.
Correct reading $= 1.51 \ mm - (-0.03 \ mm) = 1.51 \ mm + 0.03 \ mm = 1.54 \ mm$.

(D) Pitch is defined as the distance moved by the screw in one complete rotation.
Given that $5$ rotations correspond to a linear movement of $2.5 \text{ mm}$.
Therefore,Pitch = $\frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}$.
The least count of a screw gauge is calculated using the formula: $\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}}$.
Substituting the values: $\text{Least Count} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$.
In scientific notation,this is $5 \times 10^{-3} \text{ mm}$.

(A) Least count $(LC)$ = $\frac{\text{pitch}}{\text{total divisions}} = \frac{0.1 \text{ mm}}{100} = 0.001 \text{ mm}$.
Zero error = $5 \times LC = 5 \times 0.001 \text{ mm} = 0.005 \text{ mm}$.
Observed reading = $\text{Main scale reading} + (\text{Circular scale division} \times LC) = 5 \text{ mm} + (50 \times 0.001 \text{ mm}) = 5.050 \text{ mm}$.
Correct reading = $\text{Observed reading} - \text{Zero error} = 5.050 \text{ mm} - 0.005 \text{ mm} = 5.045 \text{ mm}$.

(C) Least count $(LC)$ = $1 \text{ MSD} - 1 \text{ VSD}$.
Given that $10 \text{ VSD} = 9 \text{ MSD}$,therefore $1 \text{ VSD} = 0.9 \text{ MSD} = 0.9 \text{ mm}$.
$LC$ = $1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm} = 0.01 \text{ cm}$.
Zero error = $+ (n \times \text{LC})$,where $n$ is the coinciding division.
Zero error = $+ (7 \times 0.01 \text{ cm}) = +0.07 \text{ cm}$.
Since the Vernier zero is shifted to the right of the main scale zero,the zero error is positive.

(C) The least count $(LC)$ of a vernier calliper is defined as $LC = 1 \text{ MSD} - 1 \text{ VSD}$.
Given that $20 \text{ VSD} = 16 \text{ MSD}$,therefore $1 \text{ VSD} = \frac{16}{20} \text{ MSD} = 0.8 \text{ MSD}$.
Substituting this into the formula,we get $LC = 1 \text{ MSD} - 0.8 \text{ MSD} = 0.2 \text{ MSD}$.
Since $1 \text{ MSD} = 1 \text{ mm} = 0.1 \text{ cm}$,the $LC = 0.2 \times 0.1 \text{ cm} = 0.02 \text{ cm}$.