Two water pipes P and Q having diameters 2 ti | vedclass

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(A) Given: Diameter of pipe $P$,$d_{p} = 2 	imes 10^{-2} \ m$,so radius $r_{p} = 10^{-2} \ m$.Diameter of pipe $Q$,$d_{q} = 4 	imes 10^{-2} \ m$,so

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$4$ times that of $Q$

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(A) Given: Diameter of pipe $P$,$d_{p} = 2 	imes 10^{-2} \ m$,so radius $r_{p} = 10^{-2} \ m$.Diameter of pipe $Q$,$d_{q} = 4 	imes 10^{-2} \ m$,so radius $r_{q} = 2 	imes 10^{-2} \ m$.According to the equation of continuity,the product of cross-sectional area and velocity is constant for an incompressible fluid: $A_{p}v_{p} = A_{q}v_{q}$.Therefore,the velocity is inversely proportional to the cross-sectional area: $\frac{v_{p}}{v_{q}} = \frac{A_{q}}{A_{p}}$.Since $A = \pi r^{2}$,we have $\frac{v_{p}}{v_{q}} = \frac{\pi r_{q}^{2}}{\pi r_{p}^{2}} = \left(\frac{r_{q}}{r_{p}}ight)^{2}$.Substituting the values: $\frac{v_{p}}{v_{q}} = \left(\frac{2 	imes 10^{-2}}{10^{-2}}ight)^{2} = (2)^{2} = 4$.Thus,$v_{p} = 4v_{q}$. The velocity of water in pipe $P$ is $4$ times that of $Q$.

Two water pipes $P$ and $Q$ having diameters $2 \times 10^{-2} \ m$ and $4 \times 10^{-2} \ m$,respectively,are joined in series with the main supply line of water. The velocity of water flowing in pipe $P$ is

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