Question

The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury = 13.6, Density of water $={10}^{3}kg/m^3,g=9.8m/s^2$ at Calcutta$.Pressure=hpg$ in usual symbols.

Answer 1

Height of the Mercury Column in a Barometer = 75 cm.

For Finding the Pressure in S.I. System, Changing 75 cm into meter.

∴ Height of the Mercury column(h) = 75 cm.

= 0.75 m.

Acceleration due to gravity(g) = $9.8m/s^2$

Density of the Water =${10}^{3}kg/m^3$.

=$1000kg/m^3$

Specific Gravity of the Mercury = 13.6

Specific Gravity of the Substance is the Ratio of the Density of the Substance to the Density of the Water.

∴ Specific Gravity = Density of the Substance /Density of the Water.

⇒ Density of the Mercury(ρ) = $1.3\times 1000$

= $1300kg/m^3$.

Now,

Using the Formula,

Pressure(P) = hρg

= $0.75\times 1300\times 9.8$

= 9555 Pa.

Hence, the Pressure is 9555 Pa.

For Finding the Pressure in C.G.S. System,

Height of the Mercury Column(h) = 75 cm.

Acceleration due to gravity(g) = $9.8m/s^2$

=$9.8\times 100cm/s^2$

=$980cm/s^2$.

Density of the Water =$1000kg/m^3$.

= $1000\times {10}^{6}g/{10}^{6}c{m}^3$.

= ${10}^6/{10}^{6}g/c{m}^3$.

= $1g/c{m}^3$

∴ Density of the Mercury(ρ) = Specific Gravity of the Mercury × Density of the Water.

=$1.3\times 1$

=$1.3g/c{m}^3$.

Now Using the Formula,

Pressure(P)= hρg

= $75\times 1.3\times 980$

= 9550 Ba.

Hence the Pressure is 9550 barye.