The heights of mercury surfaces in the tow arms of the manometer shown in figure are 2 cm and 8cm. Atmospheric pressure `=1.01xx10^5Nm^-2` . Find a. the pressure of the gas in the cylinder and b. the pressure of mercury at the bottom of the U tube.

Answer

A, B

Explanation

a. Given `h_1=8cm`
`h_2=2cm`
`rho_(Hg)=13.6gm/cc`
`rho_a=1.01xx10^5N/m^2`
`=1.01xx10^5xx10dyne/cm^2`
Pressure at the bottom of the tube shold be same when considered for both limbs from the ure are shown.
`rarrP_g+rho_(Hg)xxh_2xxg`
`=P_a+rho_(Hg)xxh_1xxg`
`-. Pg=P_0+rho_(Hg)xxg(h_1-h_2)`
`=1.01xx10^5+13.6xx980`
`xx(8-2)dyne/cm^2`
`=(1.01xx10^6+13.6xx980xx6)Dyne/cm^2`
`=1.09xx10^5N/m^2`
b. m=45kg` (From the ure shown)
`P^3+hpg=P_q+(mg)/A` ltbr `rarr hpg=(mg)/A`

`rarr h=m/(Ap)`
`= 45/(900xx10^-4xx10^3)`
`=0.5m=50cm