Question

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:

• A. 4 m/s
• B. 6 m/s
• C. 8 m/s
• D. 2 m/s

Hint:

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.

Answer 1

The Correct Option is C

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 \( \boxed{8} \, \text{m/s} \).