Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm$$^2$$. The area of cross section at the outlet is 10 mm$$^2$$. If the piston is pushed at a speed of 4 cm s$$^{-1}$$, the speed of outgoing fluid is _______ cm s$$^{-1}$$

We need to find the velocity of the liquid at the outlet of a tube when a piston pushes it through. We begin by noting that $$A_1 = 2.0 \text{ cm}^2$$ is the inlet area, $$A_2 = 10 \text{ mm}^2$$ is the outlet area, and $$v_1 = 4 \text{ cm/s}$$ is the piston speed.

Next, we convert $$A_2$$ to the same units as the inlet by writing $$A_2 = 10 \text{ mm}^2 = 10 \times (0.1 \text{ cm})^2 = 10 \times 0.01 \text{ cm}^2 = 0.10 \text{ cm}^2$$.

Then, applying the equation of continuity for an incompressible fluid yields $$A_1 v_1 = A_2 v_2$$.
This equality follows from conservation of mass, since the fluid entering per unit time must equal the fluid leaving per unit time.

Finally, solving for $$v_2$$ gives $$v_2 = \frac{A_1 v_1}{A_2} = \frac{2.0 \times 4}{0.10} = \frac{8.0}{0.10} = 80 \text{ cm/s}$$. Therefore, the speed of the liquid at the outlet is 80 cm/s.