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A water tank kept on the ground has an orifice of $2 \mathrm{~mm}$ diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent? (Assume, $g=10 \mathrm{~m} / \mathrm{s}^2, \rho_{\text {water }}=10^3 \mathrm{~kg} / \mathrm{m}^3$, viscosity $=1$ centi-poise)
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Verified Answer
The correct answer is:
$11 \mathrm{~cm}$
Here, $D=2 \mathrm{~mm}, \eta=1$ centi-poise $=10^{-3} \mathrm{~Pa}-\mathrm{s}$ and density of the water, $\rho=10^3 \mathrm{~kg} / \mathrm{m}^3$
For flow to be just turbulent, $R_e=3000$
$$
\therefore \quad v=\frac{R_e \eta}{\rho D}=\frac{3000 \times 10^{-3}}{10^3 \times 2 \times 10^{-3}}=1.5
$$
We know that the velocity head, $h=\frac{v^2}{2 g}$
$$
\Rightarrow \quad h=\frac{(1.5)^2}{2 \times 10}=0.1125 \simeq 11 \mathrm{~cm}
$$
So, no option is matched.
For flow to be just turbulent, $R_e=3000$
$$
\therefore \quad v=\frac{R_e \eta}{\rho D}=\frac{3000 \times 10^{-3}}{10^3 \times 2 \times 10^{-3}}=1.5
$$
We know that the velocity head, $h=\frac{v^2}{2 g}$
$$
\Rightarrow \quad h=\frac{(1.5)^2}{2 \times 10}=0.1125 \simeq 11 \mathrm{~cm}
$$
So, no option is matched.
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