1. Use equation of continuity

For an incompressible liquid, $A_1 v_1 = A_2 v_2$ where:

• $A_1$ = cross-sectional area of piston
• $v_1$ = speed of piston
• $A_2$ = outlet area
• $v_2$ = speed of outgoing fluid

2. Convert areas into same units

Given: $A_1 = 2.0\,\text{cm}^2$ $A_2 = 10\,\text{mm}^2$

Now, $1\,\text{cm}^2 = 100\,\text{mm}^2$ So, $2.0\,\text{cm}^2 = 200\,\text{mm}^2$

Thus, $A_1 = 200\,\text{mm}^2, \qquad A_2 = 10\,\text{mm}^2$

3. Substitute into continuity equation

Given piston speed: $v_1 = 4\,\text{cm s}^{-1}$

So, $A_1 v_1 = A_2 v_2$ $200 \times 4 = 10 \times v_2$ $800 = 10v_2$ $v_2 = 80\,\text{cm s}^{-1}$

4. Final answer

The speed of outgoing fluid is $\boxed{80\,\text{cm s}^{-1}}$