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Home  >>  EAMCET  >>  Physics

Water from a tap emerges vertically downwards with initial velocity $4\;ms^{-1}$. The cross- sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance h vertically below the tap, where the cross- sectional area of the stream becomes $\bigg(\large\frac{2}{3}\bigg)A,$ is $(g=10\; m/s^2)$:

\[\begin {array} {1 1} (1)\;0.5\;m & \quad (2)\;1\;m \\ (3)\;1.5\;m & \quad (4)\;2.2\;m \end {array}\]

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(2) 1 m
answered Nov 7, 2013 by pady_1
 

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