Question

A manometer tube contains mercury of density $13.6 \times {10^3}kg{m^3}$ what is difference in the levels of mercury in the two arms is indicated by a gauge pressure of $1.03 \times {10^5}Pa$ ?

Answer 1

Hint: In order to solve this question we need to understand pressure by liquids. Consider a point which is at some height and water density the pressure is defined as a product of density, height and acceleration due to gravity, as water being pulled downwards by gravity. Since pressure of a liquid only depends on height, for two holes at same height there would be equal pressure and hence the velocity of efflux only depends on height above holes rather than density of liquid.

Complete step by step answer:
According to the question, the density of liquid is given as, $\rho = 13.6 \times {10^3}kg{m^3}$.
And the pressure measured by gauge pressure is, $P = 1.03 \times {10^5}Pa$.
Let the height difference between two levels of mercury be, $h$.
Since the value of acceleration due to gravity is, $g = 9.8\,m\,{\sec ^{ - 2}}$.
So the pressure is defined as, $P = \rho gh$.
The height difference is, $h = \dfrac{P}{{\rho g}}$.
Putting values we get,
$h = \dfrac{{1.03 \times {{10}^5}Pa}}{{(13.6 \times {{10}^3}kg{m^3})(9.8m{{\sec }^{ - 2}})}}$
$\therefore h = 0.772\,m$

So the height difference between two levels of mercury is, $h = 0.772\,m$.

Note: It should be remembered that a manometer tube is a pressure measuring device using liquid columns in vertical or inclined tubes. The most common manometer is u-tube manometer, which is in the shape of u consisting of liquids, since the liquids surface level in two arms is different so a pressure exerts on them. This pressure is measured using gauge pressure. Gauge pressure is the pressure relative to atmospheric pressure as absolute pressure is defined as the sum of gauge pressure and atmospheric pressure.