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12.Kinetic Theory of Gases
normal
An open and wide glass tube is immersed vertically in mercury in such a way that length $0.05\,\, m$ extends above mercury level. The open end of the tube is closed and the tube is raised further by $0.43 \,\,m$. The length of air column above mercury level in the tube will be ...... $m$ Take $P_{atm} = 76 \,\,cm$ of mercury
A
$0.215$
B
$0.2$
C
$0.1$
D
$0.4$
Solution
Initially
$P_{i}=76 \mathrm{cm}$ of $\mathrm{Hg}$
$V_{i}=5 A$
Finally
$P_{f}=(76)-(48-x)=(28+x) \mathrm{cm}$ of $\mathrm{Hg}$
$V_{f}=x A$
since, temperature remains constant
$\therefore P_{i} V_{i}=P_{f} V_{f}$
$\Rightarrow 76 \times 5 A=(28+x) \times A$
$\Rightarrow x^{2}+28 x-380=0$
$\Rightarrow(x+38)(x-10)=0$
$\Rightarrow x=10 \mathrm{cm}$ or $x=-38 \mathrm{cm}$
$x=-38$ rejected since $x$ can't $be -ve$
Standard 11
Physics
