As per the question,
Diameter of the tube (d)=3.0cm=3×10−2m
so radius (r)=23×10−2m
mercury pressure at the bottom of the tube (P)=?
The atmospheric pressure(ρgh)=50kPa
Pressure at the bottom of the tube P=Patm+ρgh
We know that,
Density of mercury (ρ)=13546kg/m3
gravitational acceleration (g)=9.8m/sec2
Bottom pressure in excess of the atmospheric pressure
50kPa=P−Patm
⇒ ρgh=50×103Pa
⇒h=9.8×1354650×103=0.37m=37cm
⇒h=37cm
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