Question

An ideal fluid flows in a non-uniform cross-sectional tube AB as shown in the figure. If \(A_{1}\) and \(A_{2}\) is the area of cross-section and \(E_{1}\) and \(E_{2}\) is the kinetic energy per unit volume of the fluid at A and B, respectively, then the ratio \(E_{1}/E_{2}\) is

• A. \(\frac{A_{1}^{2}}{A_{2}^{2}}\)
• B. \(\frac{A_{2}^{2}}{A_{1}^{2}}\)
• C. \(\frac{A_{1}}{A_{2}}\)
• D. \(\frac{A_{2}}{A_{1}}\)

Hint:

Kinetic energy per unit volume is half density times speed squared, and the speed follows the continuity equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Kinetic energy per unit volume of a moving fluid is \(E=\dfrac{1}{2}\rho v^{2}\), so \(\dfrac{E_1}{E_2}=\dfrac{v_1^{2}}{v_2^{2}}\).

Step 2: Use continuity
For an ideal incompressible fluid, \(A_1v_1=A_2v_2\), which gives \(\dfrac{v_1}{v_2}=\dfrac{A_2}{A_1}\).

Step 3: Calculation
\(\dfrac{E_1}{E_2}=\left(\dfrac{A_2}{A_1}\right)^{2}=\dfrac{A_2^{2}}{A_1^{2}}\).
The heights at A and B only change the potential-energy term, not this ratio.

Final Answer:
The ratio \(E_1/E_2\) equals \(\dfrac{A_2^{2}}{A_1^{2}}\). \[ \boxed{\text{(B) }\dfrac{A_{2}^{2}}{A_{1}^{2}}} \]