Question

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the water go 15 m high, what fraction of the area of the hole do you have to block off with your thumb?A : 93%B : 14%C : 7.0%D : 73%E : 27%

Answer 1

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%