Air is blown through a pipe AB at the rate of 15 L/min. The cross sectional area of the wide portion of the pipe AB is cm2 and that of narrow portion is 0.5cm2 . the difference in water level is ρair =1⋅3kg/m3

Hero a1v1=A2v2=151000×60m3/s =14000m3/s now v1=14000×A1=−14000×2×10−4=1.25m/s similarly v2=14000×A2=14000×0⋅5×10−4 =5m/s Further, ρgh=12ρ1v22−v12 h=ρ12ρgv22−v12 =1⋅32×1000×9⋅8(5)2−(1⋅25)2 =1⋅55×10−3m=1⋅55mm

Air is blown through a pipe AB at the rate of 15 L/min. The cross sectional area of the wide portion of the pipe AB is cm2 and that of narrow portion is 0.5cm2 . the difference in water level is ρair =1⋅3kg/m3

Hero a1v1=A2v2=151000×60m3/s =14000m3/s now v1=14000×A1=−14000×2×10−4=1.25m/s similarly v2=14000×A2=14000×0⋅5×10−4 =5m/s Further, ρgh=12ρ1v22−v12 h=ρ12ρgv22−v12 =1⋅32×1000×9⋅8(5)2−(1⋅25)2 =1⋅55×10−3m=1⋅55mm

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Air is blown through a pipe AB at the rate of 15 L/min. The cross sectional area of the wide portion of the pipe AB is cm² and that of narrow portion is 0.5cm² . the difference in water level is $(ρ_{air} = 1 \cdot 3 kg / m^{3})$

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1.55 mm

16 mm

10 mm

3.2 mm

answer is B.

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Detailed Solution

Hero $a_{1} v_{1} = A_{2} v_{2} = \frac{15}{1000 \times 60} m^{3} / s = \frac{1}{4000} m^{3} / s now \begin{aligned} v_{1} = \frac{1}{4000 \times A_{1}} = - \frac{1}{4000 \times (2 \times 10^{- 4})} \\ = 1 .25 m / s \end{aligned} similarly v_{2} = \frac{1}{4000 \times A_{2}} = \frac{1}{4000 \times (0 \cdot 5 \times 10^{- 4})} = 5 m / s Further, ρgh = \frac{1}{2} ρ_{1} (v_{2}^{2} - v_{1}^{2}) h = \frac{ρ_{1}}{2 ρg} (v_{2}^{2} - v_{1}^{2}) = \frac{1 \cdot 3}{2 \times 1000 \times 9 \cdot 8} [(5)^{2} - (1 \cdot 25)^{2}] = 1 \cdot 55 \times 10^{- 3} m = 1 \cdot 55 mm$