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 (23)A, is (g= 10m/s2 )
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a
0.5 m
b
1 m
c
1.5 m
d
2.2 m
answer is B.
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Detailed Solution
The equation of continuity A1v1=A2v2 A×4=23A×v2 v2=6 ms−1 From Bernoulli’s theorem p+ρg1+12ρv12=p+ρgh2+12ρv22 Or g(h1−h2)=12(v22−v12) g×h=12[(6)2−(4)2] [∵ h1−h2=h] 10×h=12[36−16] h=2020=1 m