Fluid Mechanics

JEE Main level, single correct, medium

67 of 192

Question 67 of 192

Water from a tap emerges vertically downwards with an initial speed of 1.0ms−11.0m{s}^{-1} . The cross −- sectional area of the tap is 10−4m2{10}^{-4}{m}^{2} . Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross −- sectional area of the stream, 0.15m0.15m below the tap would be:
(Take g=10ms−2g=10m{s}^{-2} )

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Answer: (D) 5×10−5m25\times {10}^{-5}{m}^{2}

Velocity of water after falling through distance 0.15m0.15m is
v2=u2+2as{v}^{2}={u}^{2}+2as
=1+2×10×0.15=1+2\times 10\times 0.15
v2=4{v}^{2}=4
v=2m/sv=2 m/s
From equation of continuity
A1v1=A2v2{A}_{1}{v}_{1}={A}_{2}{v}_{2}
A2=10−4×12=5×10−5m2{A}_{2}=\dfrac{{10}^{-4}\times 1}{2}=5\times {10}^{-5}{m}^{2}

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