The mercury manometer consists of two unequal arms of equal cross section `1 cm^(2)` and lengths `100cm` and `50cm`. The two open ends are sealed with air in the tube at a pressure of `80cm` of mercury. Some amount of mercury is now introduced in the manometer through the stopcock connected to it. If mercury rises in the shorter tube to a length `10cm` in steady state, find the length of the mercury column risen in the longer tube.

Explanation

Let `p_(1)` and `p_(2)` be the pressures in `cm` of mercury in the two arms after introducing mercury in the tube. Let height of mercury column in longer tube `l_(0) cm`. using `p_(V) = const ant`.
Shorter tube: `80 xx 50 = p_(1) (50 - 10) rArr p_(1) = 100 cm`
Longer tube : `80 xx 100 = p_(2) (100 - l_(0))` (i)

From figure
`p_(1) = p_(2) + (l_(0) - 10)` (ii)
`100 = p_(2) + l_(0) - 10 rArr p_(2) = 110 - l_(0)` in (iii)
`80 xx 100 = (100 - l_(0)) (100 - l_(0))`
`= 110 xx 100 + l_(0)^(2) - 210 l_(0)`
`l_(0)^(2) - 210 l_(0) + 3000 = 0`
Solving `l_(0) = 15.5 cm`