Two water pipes of diameter 2 ,cm and 4 ,cm a | vedclass

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(A) Given: Diameter of the first pipe,$d_1 = 2 \,cm$. Radius $r_1 = 1 \,cm = 10^{-2} \,m$.Diameter of the second pipe,$d_2 = 4 \,cm$. Radius $r_2 = 2

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$4$ times that in the other pipe

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(A) Given: Diameter of the first pipe,$d_1 = 2 \,cm$. Radius $r_1 = 1 \,cm = 10^{-2} \,m$.Diameter of the second pipe,$d_2 = 4 \,cm$. Radius $r_2 = 2 \,cm = 2 	imes 10^{-2} \,m$.According to the equation of continuity for an incompressible fluid,the volume flow rate remains constant: $A_1 v_1 = A_2 v_2$.Here,$A = \pi r^2$ is the cross-sectional area.Substituting the values: $\pi (r_1)^2 v_1 = \pi (r_2)^2 v_2$.$(10^{-2})^2 v_1 = (2 	imes 10^{-2})^2 v_2$.$10^{-4} v_1 = 4 	imes 10^{-4} v_2$.$v_1 = 4 v_2$.Therefore,the velocity of flow in the $2 \,cm$ pipe is $4$ times the velocity in the $4 \,cm$ pipe.

Two water pipes of diameter $2 \,cm$ and $4 \,cm$ are separately connected to a main supply line. The velocity of flow of water in the pipe of $2 \,cm$ diameter is

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