Question

The cylindrical tube of a spray pump has a cross section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min⁻¹, what is the speed of ejection of the liquid through the holes?

Answer 1

from law of continuity ,
$A_1v_1=A_2v_2$
where $A_1$ is the cross sectional area of spray pump, $A_2$ is the cross sectional area of end of 40 holes , $v_1$ is the speed of flow of fluid inside the tube and $v_2$ is the speed of flow of the fluid through the holes.

given,
$A_1=8cm^2=8\times10^{-4}m^2$
$A_2=40\pi r^2$
where r is the radius of hole
r = 1mm/2 = 0.5mm = 5 × 10^-4 m
so,$A_2=40\times3.14\times(5\times10^{-4})^2$
$A_2=31.41\times10^{-6}$ m/s

velocity of the liquid flow inside the tube is $v_1=1.5m/min$
$v_1$ = 1.5m/60s = 0.025m/s

now, $v_2=\frac{A_1v_1}{A_2}$
= (8 × 10^-4 × 0.025)/31.41 × 10^-6
= 0.63 m/s

Therefore, the speed of ejection of the liquid through the holes is 0.63 m/s.

Answer 2

Hii dear,

◆ Answer-
v' = 0.6369 m/s

◆ Explaination-
# Given-
A = 8 cm^2 = 8×10^-4 m^2
v = 1.5 m/min = 1/40 m/s
r = 0.5 mm = 0.5×10^-3 m

# Solution-
Total area of holes-
A' = 40 × Area of each hole
A' = 40×πr^2
A' = 40×3.14×0.25×10^-6
A' = 3.14×10^-5 m^2

By equation of continuity,
Av = A'v'
v' = Av / A'
v' = (8×10^-4×1/40) / (3.14×10^-5)
v' = 0.6369 m/s

Speed of ejection of liquid through holes is 0.6369 m/s

Hope that is useful...