Question

An open U -tube of uniform cross-section contains mercury. When 27.2 cm of water is poured into one limb of the tube, (a) how high does the mercury rise in the limb from its initial level?
(b) what is the difference in levels of liquids of the two sides?
(\[{{\rho }_{w}}\]=1 and \[{{\rho }_{hg}}\]=13.6 units)

Answer 1

Hint: These works on the principle of Bernoulli’s theorem. Initially there was only mercury in the U tube and as such the height of the mercury must be the same in both the arms of the u tube. After we pour in some water, since mercury and water do not get mixed up, the mercury rises in one of the columns.

Complete step by step solution:
(a) Both the arms are open and exposed to the atmosphere, so pressure at the ends must be the same and equal to atmospheric pressure.
From Bernoulli’s theorem: \[P+\dfrac{\rho {{v}^{2}}}{2}+\rho gh\]
Since no motion, the velocity for both is zero. Thus, equating for mercury and the water we get,
$\Rightarrow P+{{\rho }_{w}}g{{h}_{w}}=P+{{\rho }_{hg}}g{{h}_{hg}} \\
\Rightarrow {{\rho }_{w}}g{{h}_{w}}={{\rho }_{hg}}g{{h}_{hg}} \\
\Rightarrow {{\rho }_{w}}{{h}_{w}}={{\rho }_{hg}}{{h}_{hg}} \\
\Rightarrow 1\times {{h}_{w}}=13.6\times {{h}_{hg}} \\
\Rightarrow 1\times 27.2=13.6\times 2x \\
\Rightarrow x=\dfrac{27.2}{13.6\times 2} \\
\therefore x=1cm \\$
Since the rise from the initial level, we have taken the height to be twice.
(b) Difference in levels of liquids of the two sides$27.2-2x=25.2cm$

Note: while solving all these kinds of problems involving the U tube, or the manometer or the hydrolift, the working theorem is Bernoulli's theorem. According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, nonviscous fluid in a streamlined flow remains a constant.
This holds for non viscous liquids and in this problem we were given water and mercury and both are non viscous.