A tube of length L is shown in the figure. The radius of cross section at 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:
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In fluid dynamics, the continuity equation for an incompressible fluid ensures that the mass flow rate is constant throughout the flow. The equation \( A_1 v_1 = A_2 v_2 \) links the velocity and cross-sectional area at different points in the tube.
Per the continuity equation for incompressible fluids, the mass flow rate is constant between any two points within a tube. The equation is:
\[A_1 v_1 = A_2 v_2\]
Here, \( A_1 \) and \( A_2 \) represent the cross-sectional areas at points (1) and (2), respectively, while \( v_1 \) and \( v_2 \) denote the velocities at these same points.
The cross-sectional area of a tube is calculated as:
\[A = \pi r^2\]
Given a radius at point (1) of \( r_1 = 2 \, \text{cm} \) and at point (2) of \( r_2 = 1 \, \text{cm} \), substituting these into the continuity equation yields:
\[\pi r_1^2 v_1 = \pi r_2^2 v_2\]
Simplification results in:
\[r_1^2 v_1 = r_2^2 v_2\]
With \( r_1 = 2 \, \text{cm}, r_2 = 1 \, \text{cm}, \) and \( v_1 = 2 \, \text{m/s} \) substituted:
\[(2^2)(2) = (1^2)(v_2)\]
\[8 = v_2\]
Consequently, the velocity of the fluid exiting point (2) is \( \boxed{8} \, \text{m/s} \).
