A constant feed of 400 mL/s is maintained by a Xanthate column of height H as shown in the figure. The outlet cross section area is \(1.0 \times 10^{-4} \, {m}^2\). The acceleration due to gravity is \(10 \, {m/s}^2\). Neglecting friction and other losses, the value of H, in cm, is (rounded off to 2 decimal places):
Show Hint
Use the Bernoulli equation and continuity equation to solve for the height of a column when the flow rate and outlet area are known.
The flow rate \( Q \) is given as \( 400 \, {mL/s} = 0.4 \, {L/s} = 4.0 \times 10^{-4} \, {m}^3/{s} \).
The outlet cross-sectional area \( A = 1.0 \times 10^{-4} \, {m}^2 \).
The acceleration due to gravity \( g = 10 \, {m/s}^2 \).
We can apply the continuity equation for the flow, which relates the velocity of the fluid, the area, and the flow rate:
\[
Q = A \times v
\]
Where \( v \) is the velocity of the fluid through the outlet.
\[
v = \frac{Q}{A} = \frac{4.0 \times 10^{-4}}{1.0 \times 10^{-4}} = 4.0 \, {m/s}
\]
Now, apply the Bernoulli equation between the surface of the Xanthate column and the outlet. Since we are neglecting friction and other losses, the equation simplifies to:
\[
\frac{v^2}{2g} = H
\]
Substitute the known values for \( v \) and \( g \):
\[
H = \frac{(4.0)^2}{2 \times 10} = \frac{16}{20} = 0.8 \, {m}
\]
Finally, converting to cm:
\[
H = 0.8 \, {m} = 80 \, {cm}
\]
Conclusion: The value of \( H \) is \( \mathbf{80.00} \, {cm} \).
