Question

An air column closed in a tube sealed at one end by a $Hg$ column having height $h$ . When the tube is placed with open end down, the height of the air column is $l_{1}$ . If the tube is turned so that its open end is at the top, the height of the air column is $l_{2} .$ What is the atmospheric pressure $\left(P_{0}\right)$ -

• A. $P_{0}=\frac{h \left(l_{1} + l_{2}\right)}{\left(l_{2} - l_{1}\right)} \, cm \, of \, Hg$
• B. $P_{0}=\frac{h \left(l_{1} - l_{2}\right)}{\left(l_{1} + l_{2}\right)} \, cm \, of \, Hg$
• C. $76 \, cm \, of \, Hg$
• D. $P_{0}=\frac{h \left(l_{1} + l_{2}\right)}{\left(l_{1} - l_{2}\right)}cm \, of \, Hg$ .

Answer 1

Answer: (D)

Let $\text{P}_{\text{0}}$ be the value at atmospheric pressure



In case I:

$P_{g a s}+h \, cm \, of \, Hg=P_{0} \, cm \, of \, Hg$

In case ii

$P_{g a s}=P_{0} \, cm \, of \, Hg+h \, cm \, of \, Hg$

By applying Boyle's law between $\left(I\right)$ & $\left(I I\right)$

$\left(P_{0} \, - \, h\right)\times l_{1}=\left(P_{0} + h\right)\times l_{2}$

or $P_{0}l_{1}- \, hl_{1}=P_{0}l_{2}+hl_{2}$

or $P_{0} \, \left(l_{1} - l_{2}\right)=h \, \left(l_{1} + l_{2}\right)$

or $P_{0}=\frac{h \left(l_{1} + l_{2}\right)}{\left(l_{1} - l_{2}\right)}$