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Water flows through a horizontal pipe of variable cross-section at the rate of $20 \mathrm{~L} \mathrm{~min}^{-1}$. What will be the velocity of water at a point where diameter is 4 cm ?
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The correct answer is:
$0.2639 \mathrm{~ms}^{-1}$
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 $\quad \frac{1}{3} \times 10^{-3}=\frac{22}{7} \times(0.02)^2 \times v$
or $\quad \mathrm{v}=\frac{7 \times 10^{-3}}{3 \times 22 \times(0.02)^2}=0.2639 \mathrm{~ms}^{-1}$
$\mathrm{V}=20 \mathrm{Lmin}^{-1}=\frac{20 \times 1000}{60 \times(100)^3} \mathrm{~m}^3 \mathrm{~s}^{-1}$
$=\frac{1}{3} \times 10^{-3} \mathrm{~m}^3 \mathrm{~s}^{-1}$
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 $\quad \frac{1}{3} \times 10^{-3}=\frac{22}{7} \times(0.02)^2 \times v$
or $\quad \mathrm{v}=\frac{7 \times 10^{-3}}{3 \times 22 \times(0.02)^2}=0.2639 \mathrm{~ms}^{-1}$
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