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  • An air column dosed in a twee sealed at one end by a column having Height <img alt="\mathrm{(H)}" s

An air column dosed in a twee sealed at one end by a \mathrm{Hg} column having Height \mathrm{(H)}
whiten the Tube is Placed with opens end down. the right of air column is \mathrm{L_{1}}
If the tube is turned so that its open end is at the top. The height of the air colum s is \mathrm{L_{2}}.
What is the atomsphere pressure \mathrm{(p_{0})}-

Option: 1

\mathrm{P_{0}=\frac{H\left(L_{1}-L_{2}\right)}{\left(L_{1}+L_{2}\right)}}


Option: 2

\mathrm{\quad P_{0}=\frac{H\left(L_{1}+L_{2}\right)}{\left(L_{1}-L_{2}\right)}}


Option: 3

\mathrm{P_{0}=\frac{\left(L_{1}-L_{2}\right)}{H\left(L_{1}+L_{2}\right)}}


Option: 4

Alone of the Above


Answers (1)

best_answer

\mathrm{In \, case\, I \rightarrow P gas +H \mathrm{~cm} at \mathrm{Hg}=P_{0} \mathrm{~cm} \, at\, \mathrm{Hg}}
\mathrm{In \: case \: II \rightarrow P_{g a s}=P_{0} \: cm \, at \, Hg +H \mathrm{~cm}\, at \, \mathrm{Hg}}

By Applying Boyle's Law between (I) & (II)

\mathrm{\left(P_{0}-H\right) \times L_{1} =\left(P_{0}+H\right) \times L_{2} }
\mathrm{\text { or } P_{0} L_{1}-H L_{1} =P_{0} L_{2} \times H L_{2} }
\mathrm{\text { or } P_{0}\left[L_{1}-L_{2}\right] =H\left[L_{1}+L_{2}\right] }

\mathrm{\text { or } \quad P_{0} =\frac{H\left[L_{1}+L_{2}\right)}{\left[L_{1}-L_{2}\right)} }

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