The tube of length L is shown in the figure. The radius of cross-section at the point (1) is 2 cm and at the point (2) is 1 cm, respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be :

We can solve this problem using the equation of continuity, which states that for an incompressible fluid, the flow rate remains constant along the stream. Mathematically,

$$A_1 V_1 = A_2 V_2$$

where:

• $A_1$ and $A_2$ are the cross-sectional areas at points (1) and (2), respectively.
• $V_1$ and $V_2$ are the velocities of water at points (1) and (2), respectively.

Substituting the values:

$$(4\pi) (2) = (\pi) V_2$$

$$V_2 = 8 \text{ m/s}$$