Question

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

Short answer

The distance below stream the faucet water is $0.283\mathrm{m}$

Step-by-step solution

Step :1 Introduction 

Diameter of the faucet 1 is $D_1=16\mathrm{mm}$

Radius of the faucet is $r_1=8\mathrm{mm}$

$=8\times {10}^{-3}\mathrm{m}$

Water flow at the speed of $v_1=1.0\mathrm{m}/{\mathrm{s}}^2$

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

Diameter of the faucet when it is narrowed $D_2=10mm$

Radius of the faucet$\text{is}{r}_2=5\mathrm{mm}$

$=5\times {10}^{-3}\mathrm{m}$

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

From the equation, continuity is

$A_1{v}_1=A_2{v}_2$

$\pi r_1^2{v}_1=\pi r_2^2{v}_2$

Speed of the water in the narrowed faucet is

$v_2=\frac{v_1{r}_1^2}{r_2^2}$

$=\frac{(1.0\mathrm{m}/\mathrm{s}){\left(8\times {10}^{-3}\mathrm{m}\right)}^2}{{\left(5\times {10}^{-3}\mathrm{m}\right)}^2}$

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

$=2.56\mathrm{m}/\mathrm{s}$

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

$\rho g{h}_1+\frac{1}{2}\rho v_1^2=\rho g{h}_2+\frac{1}{2}\rho v_2^2$

Step :2 Substituution

$\rho g\left(h_1-h_2\right)=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$

$g\left(h_1-h_2\right)=\frac{1}{2}\left(v_2^2-v_1^2\right)$

$h_1-h_2=\frac{1}{2\times 9.8}\left((2.56\mathrm{m}/\mathrm{s}{)}^2-(1.0\mathrm{m}/\mathrm{s}{)}^2\right)$

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

$\rho g=0.283\mathrm{m}$