Velocity of water after falling through distance 0.15m is
v2=u2+2as
=1+2×10×0.15
v2=4
v=2m/s
From equation of continuity
A1v1=A2v2
A2=210−4×1=5×10−5m2
JEE Main 2019 — Physics Mechanics
Water from a tap emerges vertically downwards with an initial speed of 1.0ms−1 . The cross − sectional area of the tap is 10−4m2 . Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross − sectional area of the stream, 0.15m below the tap would be:
(Take g=10ms−2 )
Held on 10 Apr 2019 · Verified 6 Jul 2026.
1×10−5m2
5×10−4m2
2×10−5m2
5×10−5m2
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