Q. 164.6( 9 Votes )

# The cylindrical t

Answer :

The law of continuity states that,

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

⇒ v_{2} = A_{1}v_{1} / A_{2} ……….(i)

Where,

‘A_{1}’ is the area of cross section of the spray pump

‘A_{2}’ is the area of cross section of the end with the 40 holes

‘v_{1}’is the speed of flow of fluid inside the tube

‘v_{2}’ is the speed of flow of the fluid through the holes

Given,

Area of cross section of spray pump, A_{1} = 8 cm^{2} = 8 × 10^{-4} m^{2}

Diameter of each hole, d = 1 mm = 1 × 10^{-3} m

Radius of each hole, r = d/2 = = 0.5 × 10^{-3} m

Velocity of the liquid flow inside the tube, v_{1} = 1.5 m min^{-1}

⇒ v_{1} = 1.5/60 m sec^{-1} = 0.025 m/s

Thus,

Area of cross section of each hole, a = πr^{2}

⇒ a = 3.14 × (0.5 × 10^{-3} m)^{2}

⇒ a = 0.785 × 10^{-6} m^{2}

Total area of cross section of 40 holes, A_{2} = 40 × a

⇒ A_{2} = 40 × 0.785 × 10^{-6} m^{2}

⇒ A_{2} = 31.41 × 10^{-6} m^{2}

Therefore, from equation (i),

⇒ v_{1 =}

⇒ 0.633 m/s

__Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.__

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