Question

A thin stream of water flows smoothly from a faucet and falls straight down. At one point the water is flowing at a speed of v1 = 1.23 m/s. At a lower point, the diameter of the stream has decreased by a factor of 0.787. What is the vertical distance h between these two points?

Answer 1

Answer:

The vertical distance between these two points is 12.28 cm.

Explanation:

Given that,

Speed of water = 1.23 m/s

Diameter at lower point d'= 0.787 d

We need to calculate the speed of water

Using continuity equation

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

$\dfrac{\pi\times d^2}{4}v_{1}=\dfrac{\pi\times d'^2}{4}v_{2}$

$v_{2}=v_{1}\times(\dfrac{d}{d'})^2$

Put the value into the formula

$v_{2}=1.23\times(\dfrac{d}{0.787d})^2$

$v_{2}=1.98\ m/s$

We need to calculate the  vertical distance h between these two points

Using Bernoulli theorem

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}$

Here, P₁ = P₂ = atmosphere pressure is same because both end is open

$\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)=\rho\times g(h_{1}-h_{2})$

$(v_{2}^2-v_{1}^2)=2g(h_{1}-h_{2})$

$2g\times\Delta h=(v_{2}^2-v_{1}^2)$

$\Delta h=\dfrac{1.98^2-1.23^2}{2\times9.8}$

$\Delta h=12.28\ cm$

Hence, The vertical distance between these two points is 12.28 cm.