First, we draw the free body diagram i.e.
When open end is uppermost
In this position. the trapped air is under a total pressure \(P_{1}\) Given as;
\(P_{1}=P_{\text {stm }}+\) pressure due to \(15 cmHg\)
\(V_{1}=A_{1} /\)
When column is inverted
In this positisn. trapped air is under a total pressure.iven as:
\(P_{s}=P_{\text {ins }}\) Prssure due to \(15 cmHg\)
\(1=13=13\)y Boyle'sliw. \(P l^{\prime}=\) constant
i.e. \(P_{1} V_{1}=P_{2} V_{2}\)
\(P_{1} A_{1} l_{1}=P_{2} A_{2} l_{2}\left(A_{1}=A_{2}\right.\) since tube is uniform)
\(P_{1} l_{1}=P_{2} I_{2} \ldots \ldots(1)\)
We substitute \(P_{1}\) and \(P_{2}\) respectively
into (1)
\(\left(P_{\text {atm }}+15\right) \times 10=\left(P_{\text {atm }}-15\right) \times 25\)
\(10 P_{\text {atm }}+150=25 P_{ atm }-375\)
\(525=15 P_{\text {at }}\)
\(P_{\text {atm }}=525 / 15=35 cmHg\)
Note: when tube or column is in the horizontal position, the trapped air is under a total pressure equal to the atmospheric pressure i.e. \(P=P_{\text {atm }}\)