Q100 / 200 · NEET 2024 (Re-NEET) · Physics

An ideal fluid is flowing in a non-uniform cross-sectional tube $X Y$ (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 : Figure

  • A
    $K_1=K_2$
  • B
    $2 K_1=K_2$
  • Correct answer
    C
    $K_1\gtK_2$
  • D
    $K_1 \lt K_2$

Solution

According to Bernoulli's principle, Kinetic energy per unit volume + Potential energy per unit volume + Pressure $=$ Constant $\frac{1}{2} \rho V^2+\rho g h+P=$ constant Apply Bernoulli's principle at point $X$ and $Y$, $P+K_1+\rho g(0)=P+K_2+\rho g(\mathrm{~h})$ $K_1=K_2+\rho g h$ $K_1\gtK_2$

Hence, the correct answer is option: 'C'.

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