Question

A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 11.3cm deep.
How far upward from its initial position does the mercury in the right arm rise?(In mm)

Answer 1

Concepts and reason

The concept required to solve the given problem is Bernoulli's equation. Initially, draw a figure from the given data. Then, use Bernoulli's equation to find the pressure equation at watermercury interface. Finally, solve this equation for height raised by mercury at right arm.

Fundamentals

The Bernoulli's equation for incompressible fluids is given by following expression. \(P_{1}+\frac{1}{2} \rho_{1} v_{1}^{2}+\rho_{1} g h_{1}=P_{2}+\frac{1}{2} \rho_{2} v_{2}^{2}+\rho_{2} g h_{2}\)

Here, \(P_{1}\) is the atmospheric pressure at the surface of fluid \(1, P_{2}\) is the atmospheric pressure at the surface of fluid 2, \(\rho_{1}\) is the density of fluid 1, \(\rho_{2}\) is the density of fluid 2 , \(\mathrm{g}\) is the acceleration due to gravity, \(h_{1}\) is the height of fluid 1 and \(h_{2}\) is the height of fluid 2 .

 

Draw the following figure from given data.

Here, \(\mathrm{h}\) is the height of the mercury raised in right arm, and \(h_{w}\) is the height of the water.

The height of the mercury raised from the level of water-mercury interface is equal to \(2 \mathrm{~h}\) because the height raised by the mercury from its initial position considered as \(\mathrm{h}\).

 

The Bernoulli's equation at water-mercury that at points \(\mathrm{A}\) and \(\mathrm{B}\) is given by following expression. \(P_{A}+\rho_{\mathrm{w}} g h_{\mathrm{w}}=P_{B}+\rho_{\mathrm{Hg}} g h_{\mathrm{Hg}}\)

Here, \(P_{\text {atm }}\) is the atmospheric pressure, \(\rho_{\mathrm{w}}\) is the density of water, \(\rho_{\mathrm{Hg}}\) is the density of mercury, \(\mathrm{g}\) is the acceleration due to gravity, \(h_{\mathrm{w}}\) is the height of water column, and \(h_{\mathrm{Hg}}\) height of mercury column. The atmospheric pressures at point \(\mathrm{A}\) and \(\mathrm{B}\) is equal because both the points as same level. \(P_{A}=P_{B}\)

Substitute \(P_{A}=P_{B}\) and \(2 \mathrm{~h}\) for \(h_{H g}\) in the above equation to solve for \(\mathrm{h} .\)

$$ \begin{array}{c} P_{A}+\rho_{\mathrm{w}} g h_{\mathrm{w}}=P_{A}+\rho_{\mathrm{Hg}} g(2 h) \\ h=\frac{\rho_{\mathrm{W}}}{2 \rho_{\mathrm{Hg}}} h_{\mathrm{w}} \end{array} $$

Substitute \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\mathrm{w}}, 13600 \mathrm{~kg} / \mathrm{m}^{3}\) for \(\rho_{\mathrm{Hg}}\), and \(11.3 \mathrm{~cm}\) for \(h_{\mathrm{w}}\) in the above equation.

$$ \begin{array}{c} h=\frac{1000 \mathrm{~kg} / \mathrm{m}^{3}}{2\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)}(11.3 \mathrm{~cm}) \\ =0.415 \mathrm{~cm}\left(\frac{10 \mathrm{~mm}}{1 \mathrm{~cm}}\right) \\ =4.15 \mathrm{~mm} \end{array} $$

The height raised by mercury at right arm is \(4.15 \mathrm{~mm}\)

since the mercury is incompressible and water and mercury are in static equilibrium, the square of the velocity terms in Bernoulli's equation is negligible.

The height raised by mercury at right arm is \(4.15 \mathrm{~mm}\).