Water from a tap emerges vertically downwards with an initial velocity V0 . Assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of the stream is half of the cross-sectional area of the stream at the tap.

Question

Water from a tap emerges vertically downwards with an initial velocity V0 . Assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of the stream is half of the cross-sectional area of the stream at the tap. (A) (V02/2 g) (B) (3 V02/2 g) (C) (2 V02/g) (D) (5 V02/2 g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-nytnnr8ulddm1inx.jpg" /> Using the 3rd equation of motion V2=u2+2gh V2=√V02 + 2 g h Using the equation of continuity A1V1=A2V2 we get (A2/A1)=(V0/√V02 + 2 g h) Required condition (A1/A2)=(1/2)=(V0/√V02 + 2 g h) ⇒ 4V02=V02+2gh h=(3 V02/2 g)

$\frac{V _{0}^{2}}{2 g}$

$\frac{3 V _{0}^{2}}{2 g}$

$\frac{2 V _{0}^{2}}{g}$

$\frac{5 V _{0}^{2}}{2 g}$

2gV02​​

2g3V02​​

g2V02​​

2g5V02​​

Using the 3rd equation of motion V2=u2+2gh V2​=V02​+2gh​ Using the equation of continuity A1​V1​=A2​V2​ we get A1​A2​​=V02​+2gh​V0​​ Required condition A2​A1​​=21​=V02​+2gh​V0​​ ⇒4V02​=V02​+2gh h=2g3V02​​

$\frac{V _{0}^{2}}{2 g}$

$\frac{3 V _{0}^{2}}{2 g}$

$\frac{2 V _{0}^{2}}{g}$

$\frac{5 V _{0}^{2}}{2 g}$