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.
According to the continuity equation for an incompressible fluid, the mass flow rate at any two points in the tube must be equal. The continuity equation is:
\[
A_1 v_1 = A_2 v_2
\]
where \( A_1 \) and \( A_2 \) are the cross-sectional areas at points (1) and (2), and \( v_1 \) and \( v_2 \) are the velocities at points (1) and (2), respectively.
The cross-sectional area of the tube is given by:
\[
A = \pi r^2
\]
Let the radius at point (1) be \( r_1 = 2 \, \text{cm} \) and at point (2) be \( r_2 = 1 \, \text{cm} \). Substituting into the continuity equation:
\[
\pi r_1^2 v_1 = \pi r_2^2 v_2
\]
Simplifying:
\[
r_1^2 v_1 = r_2^2 v_2
\]
Substituting \( r_1 = 2 \, \text{cm}, r_2 = 1 \, \text{cm}, v_1 = 2 \, \text{m/s} \):
\[
(2^2)(2) = (1^2)(v_2)
\]
\[
8 = v_2
\]
Thus, the velocity of water leaving point (2) is \( \boxed{8} \, \text{m/s} \).
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Approach Solution -2
Step 1: Use the principle of continuity. For an incompressible fluid, the volume flow rate is constant along the tube: \[ A_1 v_1 = A_2 v_2, \] where \( A \) = cross-sectional area, \( v \) = velocity.
Step 2: Express the area in terms of radius. \[ A = \pi r^2. \] Hence, \[ \pi r_1^2 v_1 = \pi r_2^2 v_2. \] Simplify: \[ r_1^2 v_1 = r_2^2 v_2. \]
