In given figure, an ideal gas is trapped between a mercury column and the closed end of a uniform vertical tube. The upper end of the tube is open to the atmosphere. Initially the lengths of the mercury column and the trapped air colume are `12 cm` and `50cm` respectively. when the tube is titted solwly in a vertical plane through an angle of `30^(@)` with horizontal then the new length of air colume is `(x)/(41)m`. Find `x`. Assuming the temperature to remains constant. `(P_(atm) = 76 cm of Hg)`

Answer

22

Explanation

Initial pressure of gas `P_(i) = P_(0) +h rho g`
final pressure of gas `P_(2) = P_(0) +h cos 60^(@) rho g`
for air column
`P_(1) V_(1) = P_(2)V_(2)`
`rArr (P_(0) +h rho g) A xx 50 = (P_(0) +rho gh cos 60^(@)) A xx x`
`x = ((76+12)50)/((76+6))=(88xx50)/(82) = (2200)/(41)cm`.