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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: [add image]
A
$K_1=K_2$
B
$2K_1=K_2$
C
$K_1>K_2$
D
$K_1<K_2$
Detailed Solution
By Bernoulli's principle along the horizontal tube, $P+K+\rho g h=\text{constant}$. From the figure, the tube rises from X to Y so $P+K_1+\rho g(0)=P+K_2+\rho g h \Rightarrow K_1=K_2+\rho g h$, hence $K_1>K_2$.