Physics - Fluid Question with Solution | TestHub

Volume of the water flowing per second, $\mathrm{V}=20{\mathrm{Lmin}}^{-1}=\frac{20\times 1000}{60\times (100{)}^3}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ $=\frac{1}{3}\times 10^{-3}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ Radius of the pipe, $\mathrm{r}=\frac{4}{2}=2\mathrm{cm}=0.02\mathrm{m}$ Area of cross-section, $\mathrm{r}=\pi {\mathrm{r}}^2=\frac{22}{7}\times (0.02{)}^2{\mathrm{m}}^2$ Let $v$ be the velocity of the flow of water at the given point. Clearly, $\mathrm{V}=\mathrm{Av}$ or $\frac{1}{3}\times 10^{-3}=\frac{22}{7}\times (0.02{)}^2\times v$ or $\mathrm{v}=\frac{7\times 10^{-3}}{3\times 22\times (0.02{)}^2}=0.2639{\mathrm{ms}}^{-1}$