Question

what will be the height of the liquid of the density 3.4 g/cm3 in the barometer at place where it is 70 cm for mercury barometer?

Answer 1

we know, for liquid barometer ,
pressure is given , P = 70cm of Hg
e.g., P = 0.70 m
density of Mercury (Hg) liquid , ρ = 13.6 g/cm³
e.g., ρ = 13.6 kg/m³
and g = 9.8 m/s²
so, P = hρg = 0.70× 13.6 × 10³ × 9.8
= 93296 Pa

now, a/c to question,
density of liquid , ρ' = 3.4 g/cm³=3.4 × 10³ kg/m³
so, P = P' = h'ρ'g
=> 93296 = h' × 3.4 × 10³ × 9.8
=> h' = 93296/(3.4 × 10³ × 9.8)
= 2.8 m

so, height = 2.8 m

Answer 2

Answer:

h=280 cm

Explanation:

we know, for liquid barometer,

pressure is given,

p=70cm of Hg

e.g....,p=0.70m

density of Mercury (Hg) liquid ,

p=13.6kg/m^3

and g=9.8m/s^2

so,

p=hpg

=0.70×13.6×10^3×9.8

p=93296pa

Now, according to question,

density of liquid,

P'=3.49/cm^3

P'=3.4 ×10^3 kg/m^3

so,

p=p'=h'p'g

93296=h'×3.4×10^3×9.8

h' =2.8m

so height is 280cm