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Chapter 14: Q.62 (page 386)

Water flowing out of a $16 m m$ -diameter faucet fills a $20 L$ bottle in $10 s$ . At what distance below the faucet has the water stream narrowed to $10 m m$ diameter?

Short Answer

Expert verified

The distance below stream the faucet water is $0.283 m$

Step by step solution

01

Step :1 Introduction

Diameter of the faucet 1 is $D_{1} = 16 mm$

Radius of the faucet is $r_{1} = 8 mm$

$= 8 脳 10^{鈭� 3} m$

Water flow at the speed of $v_{1} = 1.0 m / s^{2}$

Acceleration due to gravity is $g = 9.80 m / s^{2}$

Diameter of the faucet when it is narrowed $D_{2} = 10 m m$

Radius of the faucet $is r_{2} = 5 mm$

$= 5 脳 10^{鈭� 3} m$

Water flow after the faucet is narrowed $v = ?$

From the equation, continuity is

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

$蟺 r_{1}^{2} v_{1} = 蟺 r_{2}^{2} v_{2}$

Speed of the water in the narrowed faucet is

$v_{2} = \frac{v_{1} r_{1}^{2}}{r_{2}^{2}}$

$= \frac{(1.0 m / s) {(8 脳 10^{鈭� 3} m)}^{2}}{{(5 脳 10^{鈭� 3} m)}^{2}}$

$= \frac{64}{25} m / s$

$= 2.56 m / s$

The pressure difference is the same in both of the faucets. Therefore, from Bernoulli's equation

$蚁 g h_{1} + \frac{1}{2} 蚁 v_{1}^{2} = 蚁 g h_{2} + \frac{1}{2} 蚁 v_{2}^{2}$

02

Step :2 Substituution

$蚁 g (h_{1} 鈭� h_{2}) = \frac{1}{2} 蚁 (v_{2}^{2} 鈭� v_{1}^{2})$

$g (h_{1} 鈭� h_{2}) = \frac{1}{2} (v_{2}^{2} 鈭� v_{1}^{2})$

$h_{1} 鈭� h_{2} = \frac{1}{2 脳 9.8} ((2.56 m / s)^{2} 鈭� (1.0 m / s)^{2})$

$= \frac{1}{19.6} (5.5536)$

$蚁 g = 0.283 m$

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