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

To solve this problem, we need to apply the principles of fluid dynamics, specifically Bernoulli’s equation, which states that for an ideal fluid in steady flow, the total energy per unit volume is constant along a streamline. Bernoulli’s equation is given by: 

\[\frac{1}{2} \rho v^2 + \rho gh + P = \text{constant}\]

where \(\rho\) is the fluid density, v is the flow velocity, g is the acceleration due to gravity, h is the height above a reference point, and P is the pressure.

Since the tube is horizontal, the gravitational potential energy term \(\rho gh\) cancels out. Thus, the equation simplifies to:

\[\frac{1}{2} \rho v_1^2 + P_1 = \frac{1}{2} \rho v_2^2 + P_2\]

Considering the kinetic energy per unit volume at points X and Y:

• At point X: \(K_1 = \frac{1}{2} \rho v_1^2\)
• At point Y: \(K_2 = \frac{1}{2} \rho v_2^2\)

Using the continuity equation, \(A_1 v_1 = A_2 v_2\), where \(A_1\) and \(A_2\) are cross-sectional areas at X and Y respectively, we can solve for the velocities.

Since the cross-sectional area decreases from X to Y, the velocity increases (i.e., \(v_2 > v_1\)). Thus, the kinetic energy per unit volume at Y will be larger.

Hence, to maintain Bernoulli's equation, P_1 must be larger than P_2, leading to:

\[2K_1 = K_2\]

Therefore, the correct answer is: \(2K_1 = K_2\).