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?
Answers
Answered by
42
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
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
Answered by
12
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
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