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?

Asked by Pragya Singh | 1 year ago |  83

##### Solution :-

Given:

Number of holes, n = 40

Cross-sectional area  of the spray pump, A= 8 cm -2

= 8 × 10-4 m-2

Radius of each hole, r = 0.5 × 10-3 m

Cross-sectional area of each hole, a = πr2

= π (0.5 × 10-3 )2 m2

Total area of 40 holes, A2= n × a

$$40 \times \pi \left ( 0.5 \times 10^{ -3 } \right )^{ 2 }$$

= 3.14 x 10-5 m2

Speed of flow of water inside the tube, V1 = 1.5 m/min = 0.025 m/s

Let, the water ejected through the holes at a speed = V2

Using the law of continuity:

A1V1 = AV2

V2 = $$\dfrac{ A_{ 1 } V_{ 1 } }{ A_{ 2 } }​​ = \dfrac{ 8 \times 10^{ -4 }\times 0.025 }{ 3.14 \times 10^{ -5 } }$$

Therefore, V= 0.636 m/s

Answered by Abhisek | 1 year ago

### Related Questions

#### According to the law of atmospheres density of air decreases with increase in height y

(a) According to the law of atmospheres density of air decreases with increase in height y as $$ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}$$. Where ρ0 =1.25 kg m-3 is the density of air at sea level and yis a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

(b) A zeppelin of volume 1500 m3 is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y0 = 8000 m and ρHe =0.18 kg m-3].

#### Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends.

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).

#### Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius $$2.0 × 10^{–5} m$$ and density $$1.2 × 10^3 kg m^{–3}$$. Take the viscosity of air at the temperature of the experiment to be $$1.8 × 10^{–5} Pa$$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
A plane is in level flight at constant speed and each of its wings has an area of $$25 m^2$$. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be $$1 kg/m^3$$), g = $$9.8 m/s^2$$