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 lenghts of the mercury column and the trapped air column are `20cm` and `43cm` 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

Boyle's law holds good because the temperature is constant so, `P_(1)V_(1) = P_(2)V_(2)`
`P_(1)Al_(1) = P_(2)Al_(2)` or `P_(1)l_(1) = P_(2)l_(2)`, (`A` is constant)
since the bore is uniform
`P_(1) = 76 + 20 = 96 cm` of `Hg, l_(1) = 43cm,`
`P_(2) = 76 + hcostheta = 76 + 20 cos 60^(0)`
`= 76 +10 = 86 cm` of `Hg, l_(2) = ?`
i.e `96 xx 43 = 861_(2) rarr l_(2) = 48cm` of `Hg`