Verified Solution • 2025–26 Session moderate

QJEEM19Phc3b0: JEE Main 2019 Physics - Class 11 JEEM19Phc3b0 Solution: Water from a tap emerges vertically downwards with an initial…

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

Multiple Choice Options

  • A 5 × 10–4 m2
  • B 2 × 10–5 m2
  • C 5 × 10–5 m2
  • D 1 × 10–5 m2

Step-by-Step Verified Solution

Using Bernoullie’s equation v2=v12+2gh{v_2} = \sqrt {v_1^2 + 2gh}

Equation of continuity
A1V1 = A2V2

(1 cm3)(1m/s) = (A2)((1)2+2×10×15100)\left( {{A_2}} \right)\left( {\sqrt {{{\left( 1 \right)}^2} + 2 \times 10 \times {{15} \over {100}}} } \right)

⇒A2(ln⁡cm2)=12\Rightarrow {A_2}\left( {\ln c{m^2}} \right) = {1 \over 2}

⇒A2=5×10−5m2\Rightarrow {A_2} = 5 \times {10^{ - 5}}{m^2}

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