JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Fluid Mechanics

Water from a tap emerges vertically downwards with an initial speed of

1.0 ms^{- 1}

. The cross-sectional area of the tap is

1 0^{- 4} m^{2}

. Assume that the pressure is constant throughout the stream of water and that flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be:
(take

g = 10 ms^{- 2}

)

(A) $5 \times 1 0^{- 4} m^{2}$
(B) $5 \times 1 0^{- 5} m^{2}$ ✓ Correct
(C) $1 \times 1 0^{- 5} m^{2}$
(D) $2 \times 1 0^{- 5} m^{2}$

Answer: (B)

Solution

By continuity the stream carries the same volume flow everywhere, so

A_{1} v_{1} = A_{2} v_{2}

. As the water falls freely through

h = 0.15 m

it speeds up, and

v_{2} = v_{1}^{2} + 2 g h = 1^{2} + 2 \times 10 \times 0.15 = 4 = 2 m s^{- 1}

. The stream must therefore narrow to

A_{2} = A_{1} \frac{v _{1}}{v _{2}} = 1 0^{- 4} \times \frac{1}{2} = 5 \times 1 0^{- 5} m^{2}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Fluid Mechanics

Solution

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