An ideal fluid is flowing in a non-uniform cross-sectional tube $X Y$ (as shown in the figure) from end $X$…
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 :
$K_1=K_2$
$2 K_1=K_2$
$K_1 \gt K_2$
$K_1 \lt K_2$
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 $Y$,
$P+K_1+\rho g(0)=P+K_2+\rho g(\mathrm{~h})$
$K_1=K_2+\rho g h$
$K_1 \gt K_2$
