Water from a tap emerges vertically downwards with initial velocity 4 ms-1. The cross-sectional areaof the tap is A. The flow is steady and pressure is constant throughout the stream of water. Thedistance h vertically below the tap, where the cross-sectional area of the stream becomes (2/3) A, is(Take, g= 10 m s-2)
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a
2 m
b
1 m
c
0.5 m
d
4 m
answer is B.
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Detailed Solution
The equation of continuity,A1v1=A2v2⇒ A×4=23A×v2⇒ v2=6ms−1From Bernoulli's theorem,p+ρgh1+12ρv12=p+ρgh2+12ρv22 or gh1−h2=12v22−v12 or g×h=12(6)2−(4)2 ∵h1−h2=hor 10×h=12[36−16]or h=2020=1m