The radius of cross-section of the cylindrical tube of a spray pump is 2 cm . One end of the pump has 50…

The radius of cross-section of the cylindrical tube of a spray pump is 2 cm . One end of the pump has 50 fine holes each of radius 0.4 mm . If the speed of flow of the liquid inside the tube is $0.04 \mathrm{~ms}^{-1}$, the speed of ejection of the liquid from the holes is
  1. $6 \mathrm{~ms}^{-1}$
  2. $2 \mathrm{~ms}^{-1}$
  3. $4 \mathrm{~ms}^{-1}$
  4. $3 \mathrm{~ms}^{-1}$

Solution

$r_1=2 \mathrm{~cm}, r_2=0.04 \mathrm{~cm}, v_1=0.04 \mathrm{~ms}^{-1}$ By equation of continuity, $A_1 v_1=50 A_2 v_2 \Rightarrow \pi(2)^2 \times 0.04=50 \times \pi(0.04)^2 \times v_2$ $\therefore \quad \mathrm{v}_2=2 \mathrm{~ms}^{-1}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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