what is the pressure exerted by 75 cm vertical column of Mercury of density 13600 kgm cube in SI unit
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Answered by
22
p=h*row*g
given: h=75 cm = 75/100m
row= 13600kg/m^3
g=9.8m/sec^2
p= 75/100*13600*9.8
= 99960pascals.
given: h=75 cm = 75/100m
row= 13600kg/m^3
g=9.8m/sec^2
p= 75/100*13600*9.8
= 99960pascals.
Answered by
9
Pressure = Force / Area
= mg / A
= (V × d × g) / A ………[∵ Mass = Volume × Density]
= (A × L × d × g) / A …………[∵ Volume = Area × Length]
= Ldg
= 0.75 m × 13600 kg/m³ × 9.8 m/s²
= 99960 Pa
= 99.96 kPa
Pressure exerted is 99.96 kPa
= mg / A
= (V × d × g) / A ………[∵ Mass = Volume × Density]
= (A × L × d × g) / A …………[∵ Volume = Area × Length]
= Ldg
= 0.75 m × 13600 kg/m³ × 9.8 m/s²
= 99960 Pa
= 99.96 kPa
Pressure exerted is 99.96 kPa
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