An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. the atmospheric pressure equals 76cm of mercury. The lengths of the mercury column and trapped air column are `20 cm` and `43 cm` respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of `60^(@)`? Assume the temperature to remain constant.

Explanation

Case 1 : Atmospheric pressure + Pressure
due to mercury column
Case 2 : Atmospheric pressure +
Component of the pressure due
to mercury column
`((P_1)(V_1) = (P_2)(V_2))`
` rArr (76 xx (rho_Hg) xx g + (rho_(Hg)) xx g xx 20 ) xx A xx 43`
` = (76 xx(rho_(Hg)) xx g + (rho_ (Hg)) xx g`
` xx 20 xx (cos 60 ^@) xx A xx 1 `
` rArr 96 xx 43 = 86 xx L `
` rArr L= (96 xx 43 / 86 ) = 48 cm. `