Two water pipes of diameters 2 text{ cm} and | vedclass

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(A) Given: Diameter of first pipe $d_A = 2 	ext{ cm}$,so radius $r_A = 1 	ext{ cm}$.Diameter of second pipe $d_B = 4 	ext{ cm}$,so radius $r_B = 2

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$4$ times that in the other pipe

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(A) Given: Diameter of first pipe $d_A = 2 	ext{ cm}$,so radius $r_A = 1 	ext{ cm}$.Diameter of second pipe $d_B = 4 	ext{ cm}$,so radius $r_B = 2 	ext{ cm}$.According to the equation of continuity,the product of the cross-sectional area and the velocity of flow is constant $(A_1v_1 = A_2v_2)$.Therefore,$\frac{v_A}{v_B} = \frac{A_B}{A_A} = \frac{\pi (r_B)^2}{\pi (r_A)^2}$.Substituting the values: $\frac{v_A}{v_B} = \left( \frac{2}{1} ight)^2 = 4$.Thus,$v_A = 4v_B$. The velocity in the $2 	ext{ cm}$ pipe is $4$ times the velocity in the $4 	ext{ cm}$ pipe.

Two water pipes of diameters $2 \text{ cm}$ and $4 \text{ cm}$ are connected with the main supply line. The velocity of flow of water in the pipe of $2 \text{ cm}$ diameter is

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