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: Understanding the Concept:
In the figure, the tube is narrow at end P (area \(A_P\)) and wider at end Q (area \(A_Q\)), and the fluid flows from P to Q. The fluid is incompressible and ideal.

Step 2: Key Formula or Approach:
1. Equation of continuity: \(A_Pv_P = A_Qv_Q\).
2. Kinetic energy per unit volume: \(K = \dfrac12\rho v^2\).

Step 3: Detailed Explanation:
Since \(A_P < A_Q\), the continuity equation gives
\[ v_P = \frac{A_Q}{A_P}\,v_Q > v_Q \]
The density \(\rho\) is the same at both ends, so \(K\) depends only on the square of the speed:
\[ K_P = \frac12\rho v_P^2 > \frac12\rho v_Q^2 = K_Q \]
So \(K_P > K_Q\). The ratio is \(\left(\dfrac{A_Q}{A_P}\right)^2\), which is greater than 1 and is not a fixed number like \(\tfrac12\). So option (A) is not a general result, and option (B) would need equal speeds, which needs equal areas.

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