Pressure due to this height of \(Hg\), acts downward on the trapped air
In this position, the trapped air is under a total pressure \(P_{1}\) \(P_{1}=P_{\text {amm }}+\) pressure due to \(15 cm\) of \(Hg\)
\(P_{1}=H+15 cm\) (since \(P_{\text {atm }}\) and \(P\) due to \(15 cm Hg\) both act \(P=H+15 cm\) downward on the trapped air)
In this position, the air is under a total pressure
\(P_{2}=P_{2 \text { atm }}-15 cm\) \(P_{2}=( H -15) cm\)
\(P_{2}=( H -15) cm\) In this position, the trapped air is under a total pressure \(P_{3}\)
\(=P_{\text {stm }}=H\)y Boyle's law, \(P V=\) constant
i.e. \(P_{1} V_{1}=P_{2} V_{2}\)
where \(V=A \times l\)
\(P_{1} A_{1} l_{1}=P_{2} \times A_{2} l_{2}\)
Since the tube is uniform, \(A_{1}=A_{2}\)
So that
\(P_{1} l_{1}=P_{2} l_{2}\)
\((H+15) \times 10=( H -15) \times 15\)
\(H+15=(H-15) \times 1.5\)
\(H+15=1.5 H-22.5\)
\(H=75 cm\)
\(P_{ atm }=75 cm\)