Question

a manometer tube contains mercury of density 13.6* 10 to the power 3 kg per metre cube What what difference in the level of mercury in the two arms is indicated by a gauge pressure of 1.03 * 10 to the power 5 pascals

Answer 1

we have to use the expression ,
$P=\rho\times\:g\times\:h$
where , P is the pressure in tube column,
ρ is the density of liquid,
g is the acceleration due to gravity,
h is the height of Mercury level.

now,Let difference in the level of Mercury in two arms in indicated by gauge pressure is ∆h

$\triangle{h} = \frac{P}{\rho\times\:g}$
here, P = 1.03 × 10^5 N/m²
ρ = 13.6 × 10³ Kg/m³ and g = 9.8 m/s²
so,∆h = 1.03 × 10^5/13.6 × 10^3 × 9.8
= 103/13.6 × 9.8 m
= 0.772 m = 77.2 cm

Answer 2

Given :

density of mercury=13.6 x 10³kg/m3
pressure=P=1.03 × 10⁵ pascals
Formula to be used :P=ρgh

wher ρ=density of liquid
g-acceleration due to gravity
h is height of liquid column

Difference of mercury in both arms:

P=ρgΔh

Δh=P/ρ g
=1.03x10⁵ pa/13.6 x 1000 x9.8
=0.772m
=77.2cm