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

Case `①`: `h_(0) + l_(0) = h_(1) rArr 76 + 20 = h_(1)`
`h_(1) = 96 cm`
Case `②`: `h_(0) + l_(0) cos 60^(@) = h_(2) rArr 76 + 20 xx (1)/(2) = h_(2)`
`h_(2) = 86 cm`
`P_(1)V_(1) = P_(2)V_(2)`
`h_(1) Al_(1) = h_(2)Al_(2)`
`96 xx 43 = 86 l rArr l = 48 cm`
Note: For explanation of symbols, consider alternative Solution: of previous problem.