Question

A well of cross-sectional area $a_w$ is connected to an inclined tube of cross-sectional area $a_t$ to form a differential pressuere gauge as shown in the figure below. When $p_1=p_2$ the common liquid level is denoted by A. When $p_1>p_2$, the liquid level in the well is depressed to B, and the level in the tube rises by l along its length such that deffirence in the tube well levels is $h_d$.
The angle of inclination $\theta$ of the tube with the horizontal is

• A. $si{n}^{-1}[\frac{l}{h_d}-\frac{a_w}{a_t}]$
• B. $si{n}^{-1}[\frac{h_d}{l}+\frac{a_t}{a_w}]$
• C. $si{n}^{-1}[\frac{h_d}{l}-\frac{a_t}{a_w}]$
• D. $si{n}^{-1}[\frac{h_d}{l}+\frac{a_w}{a_t}]$

Answer 1

The correct option is C $si{n}^{-1}[\frac{h_d}{l}-\frac{a_t}{a_w}]$

By the application of preassure $P_1$ on the well, the depression is observed from point A to B and the liquid rises in the tube.

Volume of liquid = Volume of liquid
depressed in raised in the tube
the well

$\left(AB\right)a_w=l\times at$

$AB=\frac{s_t}{a_w}l$

In the triangle OMZ,

OM = AB (as can be observed from the geometry clearly)

$\therefore OM=\frac{a_t}{a_w}l$

$sin\theta =\frac{Prependicular}{Hypotenuse}=\frac{OM}{y}$

$\therefore y=\frac{OM}{sin\theta }=\frac{a_t}{a_w}=l\left(sin\theta \right)$

In triangle ZYX,
$ZX=y+l=\frac{a_{t}l}{a_{w}sin\theta }+l$

$sin\theta =\frac{XY}{ZX}=\frac{h_d}{ZX}$

$\therefore h_d=sin\theta \left(ZX\right)$

$=sin\theta [\frac{a_{t}l}{a_{w}sin\theta }+l]=\frac{a_{t}l}{a_w}+lsin\theta$

$sin\theta =\frac{h_d}{l}-\frac{a_t}{a_w}$

$\therefore \theta =si{n}^{-1}[\frac{h_d}{l}-\frac{a_t}{a_w}]$