Q. Water from a tap emerges vertically downwards with an initial speed of $1.0\ {ms}^{- 1}$. The cross-sectional area of the tap is $10^{- 4}\ m^{2}$. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is
A) $1.0 \times 10^{- 5}m^{2}$
B) $2 \times 10^{- 5}m^{2}$
C) $5 \times 10^{- 5}m^{2}$
D) $5 \times 10^{- 4}m^{2}$
Explanation
(c) Velocity of water 0.15 m below the tap is given by
$v_{2}^{2} = v_{1}^{2} + 2gh$
=${\ (1.0)}^{2} + 2 \times 10 \times 0.15$
= 1+3=4
⟹ $v_{2} = 2{ms}^{- 1}$
Now using equation of continuity, we have
$a_{1}v_{1} = a_{2}v_{2}$
$a_{2 =}\ \frac{a_{1}v_{1}}{v_{2}} = \ \frac{10^{- 4} \times 1}{2} = 5 \times 10^{- 5}m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.