Q154 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
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$\dfrac{A_1^2}{A_2^2}$
- Correct answerB$\dfrac{A_2^2}{A_1^2}$
- C$\dfrac{A_1}{A_2}$
- D$\dfrac{A_2}{A_1}$
Solution
From the equation of continuity for an incompressible ideal fluid, the volume flow rate is constant:
$A_1 v_1 = A_2 v_2$
$\Rightarrow \dfrac{v_1}{v_2} = \dfrac{A_2}{A_1}$
The kinetic energy per unit volume of the fluid is given by $E = \dfrac{1}{2} \rho v^2$, where $\rho$ is the density of the fluid.
The ratio of kinetic energy per unit volume at points A and B is:
$\dfrac{E_1}{E_2} = \dfrac{\dfrac{1}{2} \rho v_1^2}{\dfrac{1}{2} \rho v_2^2} = \left(\dfrac{v_1}{v_2}\right)^2$
Substituting the ratio of velocities from the continuity equation:
$\dfrac{E_1}{E_2} = \left(\dfrac{A_2}{A_1}\right)^2 = \dfrac{A_2^2}{A_1^2}$
Answer: $\dfrac{A_2^2}{A_1^2}$
$A_1 v_1 = A_2 v_2$
$\Rightarrow \dfrac{v_1}{v_2} = \dfrac{A_2}{A_1}$
The kinetic energy per unit volume of the fluid is given by $E = \dfrac{1}{2} \rho v^2$, where $\rho$ is the density of the fluid.
The ratio of kinetic energy per unit volume at points A and B is:
$\dfrac{E_1}{E_2} = \dfrac{\dfrac{1}{2} \rho v_1^2}{\dfrac{1}{2} \rho v_2^2} = \left(\dfrac{v_1}{v_2}\right)^2$
Substituting the ratio of velocities from the continuity equation:
$\dfrac{E_1}{E_2} = \left(\dfrac{A_2}{A_1}\right)^2 = \dfrac{A_2^2}{A_1^2}$
Answer: $\dfrac{A_2^2}{A_1^2}$
Hence, the correct answer is option: 'B'.