A thin tube of uniform cross section is sealed at both ends. It lies horizontally, the middle `5cm` containing mercury and the parts on its two sides containing air at the same pressure `p`. When the tube is held at and angle of `60^(@)` with the vertical, the length of the air column above and below the mercury pellet are `46cm` and `44.5cm` respectively. Calculate the pressure in centimeters of mercury, The temperature of the system is kept at `30^(@)C`.

Answer

C, D

Explanation

Let A be the area of cross-section of the tube.
Since temperature is the same, applying Boyle's law on the
side AB
`Pxx(xxxA)=P_2xx(X_2xxA)….(i)`
Applying Boyle's law in section CD
`Pxx(xxxA)=P_1xx(x_1xxA)…..(ii)`
From (i) and (ii)
`P_1xx(x_1xxA)=P_2xx(X_2xxA)`
`rArrP_1x_1=P_2x_2`
where `P_2=P_1+Pressure due to mercury column`

Pressure due to mercury column
`P=F/A=(mgsin 30^@)/A=(Vdgsin 30^@)/A`
`=((Axx5)xxdg sin30^@)/A=5sin30^@cm of Hg`
`P_2=P_1+5sin30^@=P_1+2.5`
Substituting this value in (iii)
`P_1xxx_1=[P_1+2.5]xxx_2`
`P_1xx46=[P_1+2.5]xx44.5`
`:. P_1=(44.5xx2.5)/1.5`
Substituting this value in (ii)
`Pxxx=(44.5xx2.5)/1.5xx46`
`rArrPxx[(46+44.5)/2]=(44.5xx2.5)/1.5xx46`
`[:' x=(x_1+x_2)/2) rArr P=75.4 cm`