A uniform tube closed at one end, contains a pellet of mercury `10cm` long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is `20cm`. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure`=75cm` of mercury.

Explanation

Let `A` : area of cross-section
`P_(0) = rho_(m)g h_(m) = rho_(m)g xx 75`
`m` : mass of mercury
`m = rho_(m) A l_(0) = rho_(m)A xx 10`
Case `①`:

`P_(1)A + mg = P_(0)A` (i)
Case `②`:

`P_(0)A + mg = P_(2)A` (ii)
Since number of moles remains same
`P_(1) V_(1) = P_(2)V_(2)`
`P_(1)Al_(1) = P_(2) Al_(2)`
`P_(1) xx 20 = P_(2) xx l` (iii)
`(P_(0) - (mg)/(A)) xx 20 = (P_(0) + (mg)/(A))l`
`(rho_(m)g xx 75 - rho_(m)g xx 10) xx 20 = (rho_(m) xx 75 + rho_(m)g xx 10) xx l`
`65xx20=85 l`
`l = 15.3 cm`
Alternatively:
Let `P_(0) = h_(0) cm` of mercury
`P_(1) = h_(1) cm` of mercury
`P_(2) = h_(2) cm` of mercury
`l_(0)` : thickness of mercury column
`l_(2)`: `20 cm, l_(2) = l cm, h_(0) = 75 cm, l_(0) = 10 cm`
Case `①`: `P_(1) + (mg)/(A) = P_(0) rArr h_(1) + l_(0) = h_(0)`
`h_(1) = h_(0) - l_(0) = 75 - 10 = 65 cm`
Case `②`: `P_(0) + (mg)/(A) = P_(2) rArr h_(0) + l_(0) = h_(2)`
`h_(2) = h_(0) + l_(0) = 75 + 10 = 85 cm`
`P_(1)V_(1) = P_(2)V_(2) rArr h_(1) Al_(1) = h_(2)Al_(2)`
`h_(1)l_(1) = h_(2)l rArr 65 xx 20 = 85 l rArr = 15.3 cm`