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# A tube sealed at both ends is 100cm long.It lies horizontally,the middle 0.1m containing mercury & 2 ends containing air at standard atmospheric pressure.If tube is turned to vertical position,what amount of mercury will be displaced?

$\begin{array}{1 1}(A)\;1cm\\(B)\;2cm\\(C)\;3cm\\(D)\;4cm\end{array}$

$P_0V_0=P_1V_1$
$\Rightarrow P_1=\large\frac{P_0V_0}{V_1}=\frac{P_0(45)}{45+x}$
$P_0V_0=P_2V_2$
$\Rightarrow P_2=\large\frac{P_045}{45-x}$
$P_2=P_1+10$
$P_0=76cm$ Hg
$\large\frac{45P_0}{45+x}$$+10=\large\frac{45P_0}{45-x}$
$\Rightarrow x^2+684x-2025=0$
$x=2.95\approx 3$
Hence (C) is the correct answer.