Answer to Question #217618 in Classical Mechanics for Buba

Question

A manometer has a wall cross-sectional area of 4 cm3 and a tube cross-sectional area of 2 cm3 the tube is inclined at an angle 30 degree celsius to the well. If the well is exposed to a source of pressure, which result in the drop of the liquid of 10cm3. Calculate;1. The height of liquid dropped in well. 2. The height of liquid raised in the tube.

Answer 1

Given,

Cross-sectional area of the wall of the manometer $(A_1)=4cm^2$

Cross-sectional area of the wall of the tube $(A_1)=2cm^2$

Angle of inclination of the tube $(\theta)=30^\circ$

Force applied by the source in the side of well = Force applied by the water due to raised in the height of liquid

$\Rightarrow$ $P\times \Delta A\sin(30^\circ)=\rho \times V\times g$

$\Rightarrow P = \frac{\rho\times V\times g}{\Delta A \sin(30^\circ)}$

Now, substituting the values,

$\Rightarrow P= \frac{1000\times 10 \times 10^{-6}\times 10}{(4-2)\times 10^{-4}}N/m^2$

$\Rightarrow P = \frac{10^{3}}{2}N/m^2$

$\Rightarrow P =500Pa$