Question

An incompressible fluid (ideal fluid) is flowing through non uniform cross-sectional tube PQ as shown in the figure from end P to end Q. If \(K_P\) and \(K_Q\) are the kinetic energy per unit volume of the fluid at end P and end Q respectively, then

• A. \(K_P = \frac{1}{2}K_Q\)
• B. \(K_P = K_Q\)
• C. \(K_P < K_Q\)
• D. \(K_P > K_Q\)

Hint:

Continuity A v = constant. The narrow end P has the larger speed, so K = (1/2) rho v^2 is larger at P.

Answer 1

The Correct Option is D

Step 1: Same volume flow rate:
The volume of fluid crossing each section per second is the same, $Q = Av$. A smaller cross-section forces a higher speed to carry the same flow.

Step 2: Compare the speeds:
The figure shows the tube widening from P to Q, so $v_P > v_Q$.

Step 3: Energy per unit volume:
Because $\frac{K}{\text{volume}} = \frac12\rho v^2$ and $\rho$ is constant, the end with higher speed has more kinetic energy per unit volume. That is end P.

Final Answer:
$K_P > K_Q$, option (D). \[ \boxed{K_P>K_Q \text{ (D)}} \]