A regular hexagon has 6 vertices. From any one vertex, we can draw diagonals to all other vertices except:

• The vertex itself
• The two adjacent vertices (these would be sides, not diagonals)

So from one vertex, we can draw diagonals to $6 - 3 = 3$ vertices.

Therefore, the maximum number of diagonals from one vertex is $3$.

The recorded value $3.50 \text{ cm}$ has a precision of $0.01 \text{ cm}$. We must find an instrument with a least count (LC) of $0.01 \text{ cm}$.

• Option A: $LC = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm}$
• Option B: $LC = \frac{1 \text{ mm}}{50} = 0.02 \text{ mm} = 0.002 \text{ cm}$
• Option C: $LC = 0.1 \text{ cm}$
• Option D: $1 \text{ MSD} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm}$. Since $10 \text{ VSD} = 9 \text{ MSD}$, $1 \text{ VSD} = 0.9 \text{ MSD}$. $LC = 1 \text{ MSD} - 1 \text{ VSD} = 0.1 \text{ MSD} = 0.01 \text{ cm}$.

Thus, Option D matches the required precision.

• $NH_{4}NO_{2} \xrightarrow{\Delta} N_{2} + 2H_{2}O$
• $Ba(N_{3})_{2} \xrightarrow{\Delta} Ba + 3N_{2}$
• $(NH_{4})_{2}Cr_{2}O_{7} \xrightarrow{\Delta} N_{2} + Cr_{2}O_{3} + 4H_{2}O$
• $(NH_{4})_{2}SO_{4}$ decomposes to release $NH_{3}$ (Ammonia) gas, not $N_{2}$.