Question

An ideal fluid is flowing in a non-uniform cross-sectional tube XY (as shown in the figure) from end X to end Y. If $K_1$ and $K_2$ are the kinetic energy per unit volume of the fluid at X and Y respectively, then the correct option is :

• A. \(K_1 = K_2\)
• B. \(2K_1 = K_2\)
• C. \(K_1 > K_2 \)
• D. \(K_1 < K_2 \)

Answer 1

The Correct Option is B

Kinetic energy per unit volume is $\frac{1}{2}\rho v^2$, where $\rho$ is the density and $v$ is the velocity.

By the equation of continuity (for an incompressible fluid), $A_1v_1 = A_2v_2$, where A is the cross-sectional area.

From X to Y, the area increases, so the velocity decreases.

Since $K \propto v^2$, $K_1 > K_2$.