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 K₁ and K₂ are the kinetic energy per unit volume of the fluid at X and Y respectively, then the correct option is: (a) K₁ = K₂ (b) 2K₁ = K₂ (c) K₁ > K₂ (d) K₁ < K₂

Answer: (c). From the figure, end X is narrow and end Y is wide: AX vY. KE per unit volume = 1/2ρ v², so K₁ > K₂ (height does not affect it). So the answer is (c).

An ideal fluid is flowing in a non-uniform cross-sectional tube XY… | NEET 2024

Q14NEET 2024Medium

Bernoulli's Equation

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:

Show answer and solution

Answer: (c)

From the figure, end $X$ is narrow and end $Y$ is wide: $A_X < A_Y$ .
Continuity $A_X v_X = A_Y v_Y$ gives $v_X > v_Y$ .
KE per unit volume $= \frac{1}{2}\rho v^2$ , so $K_1 > K_2$ (height does not affect it).
So the answer is (c).

Report a mistake

Bernoulli's Equation question from NEET 2024

More questions on Bernoulli's Equation