Verified Solution • 2025–26 Session moderate

QNEET-2024-Ph-167: NEET 2024 Physics - Class 11 NEET-2024-Ph-167 Solution: An ideal fluid is flowing in a non-uniform cross-sectional tube …

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 K1K_1 and K2K_2 are the kinetic energy per unit volume of the fluid at X and Y respectively then the correct option is: K1=K2K_1 = K_2 2K1=K22K_1 = K_2 K1>K2K_1 > K_2 K1<K2K_1 < K_2

Multiple Choice Options

  • A K1=K2K_1=K_2 K 1 _ = K 2 _
  • B 2K1=K22 K_1=K_2 2 K 1 _ = K 2 _
  • C K_1&gt;K_2 K 1 _ > K 2 _
  • D K_1 &lt; K_2 K 1 _ < K 2 _

Step-by-Step Verified Solution

According to Bernoulli's principle, Kinetic energy per unit volume + Potential energy per unit volume + Pressure = Constant 1 2 _ V 2 + _ g h + P =  constant  Apply Bernoulli's principle at point X and Y , P + K 1 + _ g ( 0 ) = P + K 2 + _ g (   h ) K 1 = K 2 + _ g h K 1 > K 2

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