The radius of the cross-section of the cylind | vedclass

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(B) Given: Radius of the tube $r_1 = 2 \,cm = 0.02 \,m$.Radius of each hole $r_2 = 0.4 \,mm = 0.0004 \,m$.Number of holes $n = 50$.Speed of liquid ins

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(B) Given: Radius of the tube $r_1 = 2 \,cm = 0.02 \,m$.Radius of each hole $r_2 = 0.4 \,mm = 0.0004 \,m$.Number of holes $n = 50$.Speed of liquid inside the tube $v_1 = 0.04 \,ms^{-1}$.According to the equation of continuity, the volume flow rate inside the tube must equal the total volume flow rate through all the holes:$A_1 v_1 = n A_2 v_2$Substituting the areas $\pi r_1^2 v_1 = n \pi r_2^2 v_2$$r_1^2 v_1 = n r_2^2 v_2$$(0.02)^2 	imes 0.04 = 50 	imes (0.0004)^2 	imes v_2$$4 	imes 10^{-4} 	imes 0.04 = 50 	imes 16 	imes 10^{-8} 	imes v_2$$1.6 	imes 10^{-5} = 800 	imes 10^{-8} 	imes v_2$$1.6 	imes 10^{-5} = 8 	imes 10^{-6} 	imes v_2$$v_2 = \frac{1.6 	imes 10^{-5}}{8 	imes 10^{-6}} = 0.2 	imes 10 = 2 \,ms^{-1}$.

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