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Q9.16: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.16 Solution: The cylindrical tube of a spray pump has a cross-section of 8.0 …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often leave the tube area in cm² or the flow speed in m/min when applying $A_1v_1=A_2v_2$, which introduces a factor of $10^4$ or $60$ error. Always convert the area to m² (1 cm²=10⁻⁴ m²) and the speed to m/s (1 m/min=1/60 m/s) before using the continuity equation.

The cylindrical tube of a spray pump has a cross-section of 8.0 cm_ 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__, what is the speed of ejection of the liquid through the holes?

Step-by-Step Verified Solution

**Answer:** 0.637 m/s **Step-by-Step Explanation:** Area of tube A1=8.0A_1 = 8.0 cm_ =8.0×10−4= 8.0 \times 10^{-4} m_. Area of one hole a=π(0.5×10−3)2=7.85×10−7a = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m_. Total hole area A2=40×7.85×10−7=3.14×10−5A_2 = 40 \times 7.85 \times 10^{-7} = 3.14 \times 10^{-5} m_. Speed in tube v1=1.5v_1 = 1.5 m/min =0.025= 0.025 m/s. By continuity: A1v1=A2v2A_1 v_1 = A_2 v_2
v2=8.0×10−4×0.0253.14×10−5=0.637 m/sv_2 = \frac{8.0 \times 10^{-4} \times 0.025}{3.14 \times 10^{-5}} = 0.637 \text{ m/s}
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All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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