By the equation of continuity:
\[
A_1 V_1 = A_2 V_2
\]
Where \( A_1 \) and \( A_2 \) are the areas of cross-section at the piston and the outlet, and \( V_1 \) and \( V_2 \) are the velocities at the piston and the outlet respectively.
Given:
\[
A_1 = 2 \, \text{cm}^2, \quad V_1 = 4 \, \text{cm/s}^{-1}, \quad A_2 = 10 \, \text{mm}^2 = 10 \times 10^{-2} \, \text{cm}^2
\]
Substituting in the continuity equation:
\[
2 \times 4 = (10 \times 10^{-2}) \times V_2
\]
Solving for \( V_2 \):
\[
V_2 = \frac{2 \times 4}{10 \times 10^{-2}} = 80 \, \text{cm/s}^{-1}
\]
Thus, the speed of the outgoing fluid is 80 cm/s\(^-1\).