When the two jaws of a Vernier caliper are brought into contact,...

The least count (LC) of the Vernier caliper is given by $\text{LC} = \dfrac{1\text{ MSD}}{\text{Total VSD}}$.

Substituting the given values, $\text{LC} = \dfrac{1\text{ mm}}{10} = 0.1\text{ mm} = 0.01\text{ cm}$.

Since the zero of the Vernier scale lies to the left of the main scale zero, the instrument has a negative zero error.

For a negative zero error, the magnitude is calculated by subtracting the coinciding Vernier division from the total number of Vernier divisions.

Zero error $= - (\text{Total VSD} - n) \times \text{LC}$, where $n$ is the coinciding division.

Given $n = 6$ and total VSD $= 10$,

Zero error $= - (10 - 6) \times 0.01\text{ cm} = - (4 \times 0.01\text{ cm}) = -0.04\text{ cm}$.

Thus, the instrument has a $0.04\text{ cm}$ negative zero error.

Answer: $0.04\text{ cm}$ negative zero error