Question

(b) the open end of a rubber tubing of a mercury manometre is placed in a fluid of density 1.2 g/cm³. the mercury in the manometre rises by 3.0 cm. what is the depth of the fluid at the end of the rubber tubing? (density of mercury = 13.6 g/cm³).

Explanation:

Step1: Equate pressures

The pressure exerted by the fluid column is equal to the pressure exerted by the mercury column. The pressure formula is $P =
ho gh$. Let the depth of the fluid be $h_f$, density of fluid $
ho_f=1.2\ g/cm^3$, density of mercury $
ho_m = 13.6\ g/cm^3$ and height of mercury column $h_m=3.0\ cm$. So, $
ho_fgh_f=
ho_mgh_m$.

Step2: Solve for $h_f$

Cancel out the gravitational - acceleration $g$ from both sides of the equation $
ho_fh_f=
ho_mh_m$. Then $h_f=\frac{
ho_mh_m}{
ho_f}$.
Substitute $
ho_m = 13.6\ g/cm^3$, $h_m = 3.0\ cm$ and $
ho_f=1.2\ g/cm^3$ into the formula: $h_f=\frac{13.6\times3.0}{1.2}$.
$h_f = 34\ cm$.

Answer:

$34\ cm$